STRENGTH  or  MATERIALS 

A  TEXT  BOOK  FOR  TECHNICAL 
AND    INDUSTRIAL   SCHOOLS 


BY 

JOHN  PAUL  fCOTTCAMP 

MEMBER  A.S.M.E. 

Head  of  Department  of  Industrial  Mechanical  Engineering 
Pratt  Institute,  Brooklyn,  N.  Y. 


FIRST    EDITION 


NEW  YORK 

JOHN  WILEY  &  SONS,  INC 

LONDON:  CHAPMAN  &  HALL,  LIMITED 
1919 


COPYRIGHT  1919, 

feY 

JOHN  PAUL  KOTTCAMP 


OP 

BRAUNWORTH    &    CO. 

BOOK   MANUFACTURERP 

BROOKLYN.    N.    V. 


PREFACE 


THIS  text  is  the  result  of  over  twelve  years7  experience 
in  teaching  the  subject  of  strength  of  materials  to  the  stu- 
dents at  Pratt  Institute.  An  attempt  has  been  made  to 
present  the  fundamental  and  underlying  principles  with  a 
minimum  of  mathematics.  These  principles  are  applied 
by  the  introduction  of  a  large  number  of  examples  which 
are  worked  out  in  detail.  It  is  only  by  tne  working  of  orig- 
inal problems  that  the  student  is  able  to  test  his  knowledge 
of  the  subject. 

Many  of  the  applications  are  made  more  particularly  to 
mechanical  lines  of  work,  and  yet  the  student  who  proposes 
to  follow  structural  work  will  find  the  text  of  great  help  to 
him. 

The  text  covers  the  application  of  strength  of  materials 
in  the  proportioning  of  beams,  columns,  shafting,  riveted 
joints,  and  in  problems  dealing  with  simple  stresses. 

Chapter  I  has  been  made  of  an  introductory  nature,  the 
desire  being  to  present  very  briefly,  but  concisely,  the  simple 
laws  of  equilibrium,  so  that  the  student  may  be  able  quickly 
and  accurately  to  analyze  the  forces  acting  in  any  machine 
or  machine  part. 

A  chapter  is  inserted  dealing  with  the  testing,  and  the 
production  of  the  more  common  materials  of  construction. 

It  has  been  the  experience  of  the  author  that  only  by  the 
assigning  of  original  problems  in  class  is  the  instructor 
able  to  test  the  grasp  of  the  subject  the  student  is  securing, 

iii 


iv  PREFACE 

and,  therefore,  recommends  the  use  of  3X5  cards  on  which 
are  placed  problems  which  the  ctudent  has  not  seen  prior  to 
entering  class. 

The  author  desires  to  extend  his  thanks  to  his  associates, 
Mr.  Frank  O.  Price,  for  the  chapter  on  Reinforced  Con- 
crete, and  Mr.  E.  H.  MacCoul,  for  checking  the  problems 
and  valuable  suggestions  relating  to  the  text. 

JOHN  P.  KOTTCAMP. 
BROOKLYN,  JUNE,  1919. 


CONTENTS 


CHAPTER  I 

PAGE 

CONDITIONS  OF  EQUILIBRIUM 1 

ART.  1.  Definitions.  ART.  2.  Moments.  ART.  3.  Resultant 
Force,  Graphical  and  Algebraic.  ART.  4.  General  Condi- 
tions of  Equilibrium  as  Applied  to  Systems  of  Coplanar  Forces. 
Problems. 


CHAPTER  II 
SIMPLE  STRESSES 9 

ART.  5.  Unit  Stress,  Kinds  of  Stresses.  ART.  6.  Elastic 
Limit,  Table  of  Elastic  Limits  of  Materials,  Permanent  Set. 
ART.  7.  Ultimate  Strength,  Factor  of  Safety,  Real  and  Appar- 
ent. Tables.  ART.  8.  Shear,  Ultimate  Shearing  Strength. 
ART.  9.  Design  Problems,  Rules  for  Solution  of  Problems 
and  Recording  of  Calculations. 

CHAPTER  III 
MATERIALS  OF  CONSTRUCTION 21 

ART.  10.  Testing  Machines,  Construction  and  Operation. 
ART.  11.  Tension  and  Compression  Tests,  Modulus  of  Elas- 
ticity. ART.  12.  Cast  Iron,  its  Manufacture  and  Properties. 
ART.  13.  Wrought  Iron,  Puddling  Furnace.  ART.  14.  Steel, 
Manufacture  and  Properties,  Open-hearth  Furnace.  ART.  15. 
Timber,  Stone,  and  Brick,  Table  of  Properties.  ART.  16. 
General  Properties  of  Construction  Materials.  ART.  17. 
Practical  Review  Problems. 


vi  CONTENTS 


CHAPTER  IV 

PAGE 

THEORY  OF  BEAMS 42 

ART.  18.  Beam  Reactions,  Method  of  Checking  Results. 
ART.  19.  Bending  Moment.  ART.  20.  General  Rule  for 
Bending  Moment  Applied  to  Various  Types  of  Beams. 
ART.  21.  Internal  Resisting  Moments,  Section  Modulus. 
ART.  22.  Center  of  Gravity,  General  Rule,  Method  of  Filing 
Calculations.  ART.  23.  Moment  of  Inertia,  Reduction  For- 
mula, Application  to  More  Common  Geometrical  Figures, 
Computations.  ART.  24.  Practical  Review  Problems. 


CHAPTER  V 
DESIGN  OF  BEAMS 70 

ART.  25.  Safe  Loads  for  Wooden  Beams,  Allowable  Fiber 
Stresses  for  Timber.  ART.  26.  Vertical  Shear,  Shear  Dia- 
grams, Method  of  Plotting.  ART.  27.  Safe  Loads  for  Steel 
Beams,  Properties  of  I  Beams.  ART.  28.  Cast-iron  Beams, 
Gear  Teeth.  ART.  29.  Beams  of  Uniform  Strength.  ART.  30. 
Modulus  of  Rupture.  ART.  31.  Practical  Problems.  4 

CHAPTER  VI 

DEFORMATIONS 93 

ART.  32.  Modulus  of  Elasticity.  ART.  33.  Deflection  of 
Beams,  General  Equations.  ART.  34.  Practical  Problems. 

CHAPTER  VII 
SHAFTING 102 

ART.  35.  Twisting  Moments,  Torque.  ART.  36.  Internal 
Resisting  Moment  in  Torsion.  ART.  37.  Polar  Moment  of 
Inertia.  ART.  38.  Shafts  in  Pure  Torsion,  Work  and  Horse- 
power. ART.  39.  Shafts  for  Combined  Torsion  and  Bending, 
Equivalent  Bending  and  Twisting  Moments.  ART.  40.  Prac- 
tical Problems,  Table  of  Horse-power  Transmitted  by  Shaft- 
ing. 


CONTENTS  vii 


CHAPTER  VIII 

PAGE 

COLUMNS 118 

ART.  41.  Analysis  of  Stresses  Acting  in  Columns.  ART.  42. 
Radius  of  Gyration.  ART.  43.  Derivation  of  the  Rankine 
Formula  for  Columns.  ART.  44.  Cast-iron  Columns.  ART. 
45.  Structural  Columns.  ART.  46.  Comparison  of  Column 
Formulae.  ART.  47.  Practical  Problems  Relating  to  Columns. 


CHAPTER  IX 

RIVETED  JOINTS 136 

ART.  48.  Thickness  of  Pipes,  Cast-iron  Water  Pipe.  ART.  49. 
Lap  Joints,  A.S.M.E.,  Standard  Rules  for  Efficiency.  ART.  50. 
Butt  Joints,  Double,  Triple  and  Quadruple  Riveted.  Effi- 
ciencies. ART.  51.  Application  of  Riveted  Joints  to  Boilers. 
ART.  52.  Practical  Problems  on  Applications  of  Riveted 
Joints,  Structural  and  Mechanical. 

CHAPTER  X 
COMBINED  STRESSES — RESILIENCE 151 

ART.  53.  Tension  and  Compression,  Crane  Hooks.  ART.  54. 
Stresses  in  Various  Machine  Frames.  ART.  55.  Modulus  of 
Resilience.  ART.  56.  Formulae  for  Elastic  Resilience.  ART.  57. 
Stresses  Due  to  Temperature.  ART.  58.  Practical  Problems. 

CHAPTER  XI 
REINFORCED  CONCRETE 169 

ART.  59.  Cement  and  Concrete.  ART.  60.  Reinforced  Con- 
crete, Types  of  Bars.  ART.  61.  Slabs  and  Beams.  ART.  62. 
Columns.  ART.  63.  Design  of  Concrete  Beams  and  Floors. 
ART.  64.  Design  of  Columns  and  Footings.  ART.  65.  Prac- 
tical Review  Problems. 


STRENGTH  OF  MATERIALS 


CHAPTER  I 
CONDITIONS  OF  EQUILIBRIUM 

ART.  I.  DEFINITIONS 

Gravity  is  the  force  by  which  all  bodies  are  attracted 
toward  the  earth's  center.  The  force  with  which  gravity 
attracts  any  body  toward  the  earth  is  called  the  weight  of 
the  body. 

Mass  is  the  quantity  of  material  in  a  body. 

Force  is  that  which  produces  or  tends  to  produce  motion 
or  change  of  motion  of  bodies.  It  manifests  itself  to  the 
feeling  by  a  tension  or  pull,  and  by  a  compression  or  push. 

The  Unit  of  Force  is  the  weight  of  a  mass  of  one  pound 
or  simply  the  pound. 

A  force  is  completely  defined  when  its  magnitude,  direc- 
tion, and  point  of  application  are  known.  In  Fig.  1  the  line 


FIG.  1. 


AB  represents  the  magnitude  of  the  force  P;    the  arrow 
located  at  the  end  of  the  line  AB  shows  that  the  force  is 


2 


STRENGTH  OF  MATERIALS 


acting  from  A  toward  B,  and  A  indicates  the  point  of  appli- 
cation of  the  force. 

The  component  of  a  force  is  the  effort  exerted  by  that 
force  in  any  given  direction  in  the 
plane   of   the   force.     In  Fig.  2  let 
the  line  OB  represent  to  a  given 
scale  the  force  P.     From  the  point 
B  drop  a  perpendicular  EM  to  the 
-X    horizontal  line  OX,  and  a  perpen- 
FlQ  2  dicular  BN  from   the  point  B  to 

the  line  0  Y.     Then  the  line  OM  will 

represent,  to  scale,  the  horizontal  component  of  the  force 
P,  and  the  line  ON  the  vertical  component  of  the  force  P. 


ART.  2.  MOMENTS 

The  moment  of  a  force  is  the  tendency  of  the  force  to  pro- 
duce rotation  of  a  body  about  any  given  point.  The  moment 
arm  is  the  perpendicular  distance  from  the  given  point  to  the 
line  of  action  of  the  force. 

The  measure  of  a  moment  is  the  product  of  the  force  in 
pounds  times  the  moment  arm  in 
feet,  hence  moments  are  expressed 
in  foot-pounds.  Moments  may  be 
expressed  in  inch-pounds  by  using 
the  force  in  pounds  and  the  moment 
arm  in  inches.  In  Fig.  3,  let  AB 
represent  a  force  of  P  pounds, 
tending  to  rotate  the  body  D  about 
the  point  0.  Draw  OC  perpendicular 
to  AB.  Then  OC,  or  p,  is  the 
moment  arm,  and  the  moment 
equals  ABXOC  =  Pp. 

In  this  text  moments  tending  to  produce  clockwise 
rotation  will  be  considered  positive,  and  those  tending  to 


FIG.  3. 


CONDITIONS  OF  EQUILIBRIUM  3 

produce    counter-clockwise    rotation    will    be    considered 
negative. 

PKOB.  1.  A  force  of  12  Ibs.  acts  at  a  distance  of  20  ins.  from  a 
given  point.  Find  the  value  of  the  moment  of  the  force;  (a)  in 
inch-pounds;  (6)  in  foot-pounds. 

PROB.  2.  The  pull  on  a  belt  is  200  Ibs.  The  radius  of  the 
pulley  is  24  ins.  Find  the  moment  of  the  belt  pull  on  the  shaft. 

PROB.  3.  Two  forces  of  50  and  75  Ibs.  act  at  distances  of 
3  ft.  and  5  ft.  from  a  given  point.  If  the  50-lb.  force  exerts  a 
positive  moment  and  the  force  of  75  Ibs.  a  negative  moment,  find 
the  resulting  moment. 


ART.  3.  RESULTANT  FORCE 

Any  single  force  which  produces  the  same  effect  on  a 
body  as  the  combined  action  of  two  or  more  forces  is  called 
the  resultant  of  those  forces.  Take,  for  example,  two  men 
pulling  on  a  vertical  hoist.  Assume  each  man  exerts  a  pull 
of  50  Ibs.  Their  combined  effort  is  100  Ibs.  A  single  pull 
of  100  Ibs.  would  then  have  the  same  effect  as  the  combined 
pull  of  the  two  men.  In  all  cases  where  the  forces  are  par- 
allel and  in  the  same  direction  the  resultant  is  easily  found  by 
adding  together  the  single  forces.  When  the  forces  make 
an  angle  with  each  other,  the  process  of  rinding  the  result- 
ant force  is  not  so  simple. 

The  resultant  of  a  system  of  forces  can  be  found  either 
graphically  or  algebraically.  The  meaning  of  resultant 
force  may  be  made  obvious  by  an  easily  performed  experi- 
ment. Secure  two  small  pulleys,  a  few  pieces  of  stout  cord, 
and  several  small  weights,  and  arrange  the  apparatus  as 
shown  in  Fig.  4.  Suppose,  for  example,  the  weight  at  the 
point  A  is  5  Ibs.,  at  point  B  3  Ibs.,  and  at  the  point  C  6  Ibs. 
Place  a  heavy  piece  of  cardboard  back  of  the  cords  and  draw 
aline  parallel  to  the  line  CA,  making  it  proportional  to  the 
weight  of  5  Ibs.,  according  to  some  assumed  scale;  in  like 


4  STRENGTH  OF  MATERIALS 

manner  draw  another  line  parallel  to  the  cord  CB  and  make 
it  proportional  to  3  Ibs.  to  the  same  scale.  Construct  the 
parallelogram  CFDEC  by  drawing  DE  parallel  to  CF  and 
DF  parallel  to  CE.  Measure  the  line  CD  and,  if  the  experi- 
ment has  been  carefully  performed,  it  will  be  found  that  the 
line  CD  will  measure  6  Ibs.  to  the  given  scale,  and  it  will 
further  be  found  that  the  line  CD  is  vertical.  Therefore  if 
the  weights  W\  and  W%  were  replaced  by  a  single  vertical 
force  acting  upward  at  the  point  C  and  equal  to  6  Ibs.,  this 


W3=31b. 


W,=51b. 


FIG.  4. 

force  would  keep  the  weight  of  6  Ibs.  from  falling.  It  is  to 
be  noted  that  the  resultant  force  is  acting  upward.  The 
force  of  6  Ibs.  acting  downward  is  called  the  equilibriant,  or 
the  force  producing  a  state  of  rest. 

By  varying  the  weights  and  changing  the  lengths  of  the 
cords  (so  as  to  secure  different  angles)  the  student  can  find 
the  resultant  for  many  other  combinations,  and  he  will 
always  find  it  to  be  represented  by  the  diagonal  of  the 
parallelogram  formed  by  the  two  forces.  From  these  simple 
experiments  the  following  law  can  be  stated: 

//  two  forces  acting  at  a  point  on  a  given  body  be  represented 


CONDITIONS  OF  EQUILIBRIUM  5 

in  magnitude  and  direction  by  the  adjacent  sides  of  a  parallel- 
ogram, the  resultant  of  these  two  forces  will  be  represented  in 
magnitude  and  direction  by  the  diagonal  of  the  parallelogram 
passing  through  this  point. 

Another  truth  demonstrated  by  the  above  experiment  is 
that  the  resultant  force  lies  in  the  same  plane  with  the  forces 
to  which  it  is  equivalent.  To  find  the  resultant  force  by  a 
graphical  method  requires  great  care  in  the  drawing  of  all 
lines  and  angles,  and  even  then  a  certain  degree  of  error  is 
introduced. 

Another  method  of  determining  the  resultant  of  two  forces 
(and  this  method  can  be  applied  where  there  are  more  than 
two  forces)  is  to  resolve  the  forces  into  their  horizontal  and 
vertical  components,  and  then  find  the  algebraic  sum  of  all 
the  horizontal  components,  indicated  by  2#,(  and  the  alge- 
braic sum  of  all  the  vertical  components,  indicated  by  SF. 
Thus  all  the  forces  are  replaced  by  a  single  vertical  force, 
and  a  single  horizontal  force  and  the  resultant  can  be  found 
from  equation, 


(1) 


PROS,  4.  Two  forces  of  80  Ibs.  and  120  Ibs.  make  an  angle  of 
30°  with  each  other.  Determine  the  resultant  by  each  of  the  three 
methods  discussed. 

PROB.  5.  The  thrust  on  the  connecting  rod  of  an  engine  is 
8000  Ibs.  Find  the  horizontal  and  vertical  components  when  the 
connecting  rod  makes  an  angle  of  20°  with  the  horizontal,  assuming 
length  of  connecting  rod  equals  five  times  the  length  of  the  crank. 

PROB.  6.  Four  forces  act  on  a  given  body  at  the  point  0.  The 
first  force  is  horizontal  and  equals  50  Ibs.  The  second  force  makes 
an  angle  of  60°  with  the  horizontal  and  equals  60  Ibs.  The  third 
force  makes  an  angle  of  30°  with  the  horizontal  and  equals  40  Ibs. 
The  fourth  force  makes  120°  with  the  horizontal  and  equals  45  Ibs. 
Find  the  magnitude  and  direction  of  the  resultant. 


6  STRENGTH  OF  MATERIALS 


ART.  4.  CONDITIONS  OF  EQUILIBRIUM 

When  any  number  of  forces  which  are  either  concurrent 
or  non-concurrent  and  lie  in  the  same  plane  tend  to  produce 
a  state  of  equilibrium,  it  has  been  shown  that  certain  general 
conditions  must  be  satisfied  by  the  forces.  The  first  of  these 
conditions  is  that  there  can  be  no  resultant  force.  The  second 
of  the  conditions  is  that  there  may  be  no  resultant  horizontal 
or  vertical  component,  and  for  this  to  be  true  the  algebraic 
sum  of  all  the  horizontal  and  the  vertical  components  must 
equal  zero.  If  the  forces  acting  on  the  body  pass  through  a 
common  point,  the  above  conditions  are  sufficient  to  insure 
a  state  of  equilibrium;  but  if  the  forces  are  non-concurrent, 
then  the  third  condition,  that  the  algebraic  sum  of  the 
moments  of  all  the  forces  about  a  given  point  must  equal 
zero,  is  also  necessary  to  determine  completely  the  state  of 
equilibrium.  A  system  of  forces  which  is  in  equilibrium 
may  produce  either  a  state  of  rest  or  a  state  of  uniform 
motion  of  the  body  on  which  the  system  acts.  Hence  the 
forces  acting  on  a  body  which  is  in  a  state  of  uniform  motion 
may  be  in  equilibrium  just  as  well  as  forces  which  act  on  a 
body  at  rest. 

In  the  design  of  a  machine  it  is  essential  that  the  forces 
acting  on  the  various  parts  of  the  machine  be  known,  and 
that  these  forces  be  in  equilibrium.  To  analyze  these  forces 
carefully  it  is  usual  to  consider  each  part  of  the  machine 
removed  and  its  place  taken  by  a  force  represented  in  mag- 
nitude and  direction  by  a  straight  line.  When  the  parts 
are  thus  replaced  by  the  forces  the  machine  is  spoken  of  as  a 
"free  body."  For  a  complete  analysis  of  the  conditions  of 
equilibrium  the  student  is  referred  to  books  on  elementary 
mechanics.  For  the  purpose  of  this  work  the  following 
general  statements  will  serve: 


CONDITIONS  OF  EQUILIBRIUM  7 

For  equilibrium  to  exist  when, 

CASE  I.     Two  forces  act  on  a  body: 

The  forces  must  be  equal  in  magnitude,  opposite  in 
direction  and  have  a  common  point  of  application. 

CASE  II.     Three  forces  act  on  a  body: 

(a)  The  forces  must  lie  in  the  same  plane. 
(6)  The  lines  of  action  of  the  forces  pass  through  a 
common  point. 

(c)  The  three  forces  may  be  represented  in  magnitude 

and  direction  by  the  sides  of  a  triangle  taken  in 
order. 

(d)  If  the  three  forces  are  parallel,  the  resultant  force 

equals  zero,  and  the  algebraic  sum  of  the 
moments  of  the  forces  about  any  given  point 
equal  zero. 

CASE  III.     Any  number  of  forces  act  on  a  body, 

(a)  If  forces  are  concurrent  and  lie  in  the. same  plane, 
they  can  be  represented  in  direction  and  mag- 
nitude by  the  sides  of  a  polygon  taken  in  order. 
(6)  Any  one  force  is  equal  and  opposite  to  the  result- 
ant of  all  the  other  forces. 

(c)  The  algebraic  sum  of  the  horizontal  and  the  ver- 

tical components  must  equal  zero. 

(d)  The  algebraic  sum  of  the  moments  of  all  forces 

about  any  given  point  must  be  zero. 

(e)  If  the  forces  are  parallel,  the  rule  (d)  under  Case 

III  may  be  applied. 

It  will  be  noted  that  in  Case  I  the  forces  are  necessarily 
concurrent.  In  Case  II  the  forces  may  be  either  concurrent 
or  parallel,  and  in  Case  III  the  forces  may  be  concurrent, 
parallel,  or  non-concurrent,  and  in  all  the  cases  the  forces  lie 
in  the  same  plane. 


8 


STRENGTH  OF  MATERIALS 


PROB.  7.  A  boiler  stay-bolt  makes  an  angle  of  25°  with  the 
shell.  If  the  steam  pressure  is  100  Ibs.  per  sq.  in.  and  the  stay 
supports  an  area  6x9  ins.  find  the  pull  acting  in  the  stay-bolt. 

PROB.  8.  In  the  locomotive  crane  shown  in  Fig.  5  the  total 
load  acting  at  the  point  A,  due  to  the  weight  of  the  bucket,  coal, 


r 


30 


FIG.  5. 


and  boom  is  4000  Ibs.    The  pull  in  the  rope  AC  is  3500  Ibs.    Find 
the  force  tending  to  crush  boom  AD. 

PROB.  9.     In  Fig.  6  let  the  weight  TF  =  1000  Ibs.     If  the  angle 
a  =30°  find  the  tension  in  the  rod  AC. 


CHAPTER  II 
SIMPLE  STRESSES 

ART.  5.     UNIT  STRESS 

THE  effect  of  an  external  force  or  load  on  a  given  part  of  a 
machine  or  structure  is  to  set  up  in  that  part  a  resistance 
which  in  every  case  is  equal  in  magnitude  and  opposite  in 
direction  to  the  given  force.  This  internal  resistance  is 
called  a  stress.  For  example,  a  rope  sustains  a  load  of  1000 
Ibs.,  the  internal  resistance  or  stress  of  the  rope  is,  therefore, 
1000  Ibs. 

A  unit  stress  is  the  resistance  offered  by  1  sq.  in.  of  the 
material.  This  unit  stress  may  be  expressed  in  pounds  per 
square  inch  (written  Ibs.  per  D"),  or  pounds  per  square  foot. 
Thus,  if  a  bar  of  steel  2  ins.  in  diameter  sustains  a  pull  of 
10,000  Ibs.,  the  unit  stress  will  equal  10,000  divided  by  the 

10  000 
area  of  the  bar,  or  — '-—-  =  3188  Ibs.  per  sq.  in. 

o.  14 

Stresses  may  be  direct  or  indirect.  The  direct  stresses 
are  tension,  compression,  and  shear,  and  are  referred  to  as 
tensile,  compressive,  or  shearing  stresses. 

The  external  loads  or  forces  producing  these  stresses  are 
referred  to  as  tensile,  compressive,  and  shearing.  The  effect 
of  a  tensile  load  is  to  produce  a  lengthening  or  elongation 
of  the  part,  resulting  in  a  reduction  of  the  cross-sectional 
area.  The  effect  of  a  compressive  load  is  to  produce  a  short- 
ening of  the  part,  or  an  increase  in  the  cross-sectional  area, 

9 


10  STRENGTH  OF  MATERIALS 

while  the  effect  of  a  shearing  load  is  to  produce  a  cutting 
of  the  parts. 

The  indirect  stresses  are  bending  or  flexure,  and  torsion  or 
twisting. 

In  direct  stresses  the  internal  resistance  is  assumed  to  be 
uniform  on  every  square  inch  of  the  material. 
Let  P  =  the  external  load  in  pounds  ; 

A  =the  cross-sectional  area  in  square  inches; 
S  =  the  unit  stress  in  pounds  per  square  inch. 

Then,  P  =  AS     .......     (2) 

Equation  (2)  is  to  be  used  when  the  area  and  the  unit 
stress  are  known. 


Equation  (3)  is  to  be  used  when  the  load  and  unit  stress 
are  known. 


Equation  (4)  is  to  be  used  when  the  load  and  area  are 
known. 

EXAMPLE:  A  round  steel  bar  sustains  a  load  of  12,000 
Ibs.  What  should  be  its  diameter  if  the  unit  stress  is  not 
to  exceed  6000  Ibs.  per  sq.  in.?  Here  the  unknown  quan- 
tity is  the  area,  therefore,  Equation  (3)  is  to  be  used,  hence 

P     12000 


The  area  of  a  circle  is  given  by  the  equation  A  =  .7854 
where  d  =  the  diameter,  or 


\. 


7854' 


SIMPLE  STRESSES  11 

The  diameter  of  the  steel  bar 


.7854 


1.6  ins. 


PROB.  10.  A  boiler  stay-bolt  is  1|  ins.  in  diameter.  If  the 
unit  stress  is  not  to  exceed  7000  Ibs.  per  sq.  in.,  find  the  load  that 
can  be  placed  on  the  bolt. 

PROB.  11.  A  wrought-iron  tension  member  in  a  bridge  sustains 
a  load  of  60,000  Ibs.  The  cross-section  of  the  member  is  2|  X4|  ins. 
Find  the  unit  tensile  stress. 

PROB.  12.  A  generator  weighing  1800  Ibs.  is  suspended  by  an 
eye-bolt  f  in.  in  diameter.  Find  the  unit  stress  in  the  shank  of  the 
bolt.  (NOTE, — Use  the  area  at  root  of  the  thread.) 


ART.  6.    ELASTIC  LIMIT. 

Whenever  an  external  load  is  applied,  causing  an  internal 
stress,  there  always  results  some  kind  of  deformation  to  the 
part  under  load.  For  example,  a  weight  attached  to  the  end 
of  a  rope  causes  it  to  stretch,  but  when  the  weight  is  removed 
the  rope  will  return  to  its  original  length.  If  the  load  be 
gradually  increased,  a  point  will  finally  be  reached  at  which, 
upon  removal  of  the  load,  it  will  be  found  that  the  rope  has 
been  permanently  increased  in  length.  This  same  principle 
can  be  applied  to  a  bar  of  steel  under  either  a  tensile  or  com- 
pressive  stress.  Hence  it  is  found  by  experiment  that  up  to 
a  certain  point  materials  can  be  externally  loaded  and  upon 
removal  of  the  load  the  materials  will  return  to  their  original 
condition.  Beyond  this  point  a  permanent  deformation  or 
set  is  found  to  remain  upon  removal  of  the  load.  Further, 
within  the  limits  of  this  point  the  amount  of  deformation 
is  found  to  be  directly  proportional  to  the  load.  For  exam- 
ple, if  a  load  of  1000  Ibs.  cause  a  bar  to  stretch  .001  of  an 
inch,  a  load  of  2000  Ibs.  may  cause  it  to  stretch  .002  of  an  inch 
The  point  at  which  the  deformation  no  longer  remains  pro- 


12 


STRENGTH  OF  MATERIALS 


portional  to  the  load  is  called  the  elastic  limit  of  the  material. 
It  therefore  follows  that  if  any  part  of  a  machine  be  loaded 
beyond  its  elastic  limit  and  the  load  be  removed,  a  per- 
manent change  of  shape  of  the  part  will  result.  Hence. in 
all  problems  in  design  no  unit  stress  is  to  be  used  which 
exceeds  the  elastic  limit  of  the  material.  Again,  it  is  found 
that  the  same  material  varies  in  its  physical  properties,  so 
that  tests  on  different  samples  of  the  same  material  may 
show  a  variation  in  the  elastic  limit.  The  elastic  limit  is 
expressed  in  pounds  per  square  inch.  Table  I  gives  the 
elastic  limit  of  the  more  common  materials  used  in  engineer- 
ing construction.  It  must  be  understood  that  these  are 
average  values  only.  For  instance,  in  the  case  of  steel  it  is 
found  that  the  elastic  limit  varies  with  the  physical  proper- 
ties of  the  steel. 

TABLE   I 

ELASTIC  LIMIT 


Material. 

Tension. 

Compression. 

Steel:  Soft 

30  000 

30  000 

Mild  

35  000 

35  000 

Hard 

55  000 

60000 

Wrought  Iron 

25  000 

25  000 

Cast  Iron  
Stone 

6,000 

20,000 
2  000 

Timber  (with  grain)  

3,200 

3,000 

PROB.  13.  In  a  tension  test  on  a  steel  rod  0.8  in.  in  diameter 
the  total  load  at  the  elastic  limit  was  18,000  Ibs.  Find  the  elastic 
limit  in  pounds  per  square  inch. 

PROB.  14.  A  stone  pier  carries  a  load  150,000  Ibs.  If  the  unit 
stress  is  one-half  the  elastic  limit,  find  the  dimensions  of  the  stone, 
assuming  its  cross-section  to  be  square. 

PROB.  15.  A  soft  steel  tie  rod  carries  a  load  of  24,000  Ibs. 
Find  its  diameter  if  the  unit  stress  equals  one-third  the  elastic  limit, 


SIMPLE  STRESSES 


13 


ART.  7.     ULTIMATE  STRENGTH 

The  maximum  load  in  pounds  per  square  inch  required 
to  rupture  a  specimen  in  tension  or  compression  is  called  the 
ultimate  strength  of  the  material.  In  the  case  of  tension 
this  is  referred  to  as  the  ultimate  tensile  strength,  and  in  the 
case  of  compression  as  the  ultimate  compressive  strength.  The 
ultimate  strength  is  determined  experimentally  (see  ART.  11) 
and  is  expressed  in  pounds  per  square  inch.  Table  II  gives 
the  ultimate  strength  in  tension  and  compression  of  the 
various  materials. 

TABLE  II 

ULTIMATE  STRENGTH 


ULTIMATE 

STRENGTH. 

Material. 

Tensile, 
Lbs.  per  Sq.  In. 

Compressive, 
Lbs.  per  Sq.  In. 

Hard  Steel  
Structural  Steel                 ... 

100,000 
60,000 

120,000 
60,000 

Wrought  Iron 

50,000 

50,000 

Cast  Iron  

20,000 

90,000 

Brass  (Cast)                     

24,000 

Timber  (with  Grain). 

10,000 

8000 

Stone 

6000 

Needless  to  state,  no  unit  stress  in  the  design  of  any 
machine  part  dare  equal  the  ultimate  strength,  for  this 
would  mean  rupture  or  failure  of  the  part.  Therefore,  the 
actual  unit  stress  must  be  considerably  below  the  ultimate 
strength  to  avoid  rupture,  and  must  be  below  the  elastic 
limit  to  avoid  permanent  deformation.  A  margin  of  safety 
must  always  be  allowed,  which  is  usually  expressed  as  a 
factor.  The  factor  of  safety  is  a  number  representing  the 
ratio  of  the  ultimate  strength  to  the  actual  unit  stress,  more 


14  STRENGTH  OF  MATERIALS 

commonly  called  the  unit  working  stress.  Expressed  as  an 
equation, 

T-,  £     £  ,         ultimate  strength 

Factor  of  safety  =  —  r-  --  ^—        —  .       .     (5) 
unit  working  stress 

or 

.,         ,  .  ultimate  strength 

unit  working  stress  =  -r-      —,  —  7——.     .     .     (6) 
factor  of  safety 

However,  there  might  be  a  factor  of  safety,  as  far  as  strength 
is  concerned,  and  still  have  a  unit  working  stress  which 
would  cause  a  permanent  deformation.  To  avoid  this 
another  term  is  introduced,  called  the  real  factor  of  safety, 
which  is  the  ratio  of  the  elastic  limit  to  the  unit  working 
stress,  or  expressed  as  an  equation, 

ry     ,  £  £     f  ,  elastic  limit 

Real  factor  of  safety  =  —  r—  —  ^-.  —        —  .       .     (7) 
unit  working  stress 

To  distinguish  these  the  factor  of  safety  based  on  ultimate 
strength  is  called  the  apparent  factor  of  safety. 

EXAMPLE.  A  steel  tie  rod  1J  ins.  in  diameter  is  subjected 
to  a  pull  of  16,000  Ibs.  Find  the  apparent  and  the  real  factor 
of  safety. 

SOLUTION.  The  area  of  a  IJ-m-  rod  =  1.22  sq.  in.  The 
unit  working  stress  equals 

16000 

-.,  00  =  13,110  Ibs.  per  sq.  in. 


The  apparent  factor  of  safety  equals  =5  (nearly). 

-Lol  lu 

35000 

The  real  factor  of  safety  equals  =  2.67. 

lollO 

Therefore  the  bar  is  safe  as  to  strength  and  as  to  permanent 
deformation. 


SIMPLE  STRESSES 


15 


The  selection  of  the  proper  factor  of  safety  and  the  deter- 
mination of  the  correct  unit  working  stress  are  largely  estab- 
lished by  experience  and  good  judgment  in  machine  design. 
Table  III  is  inserted  for  the  convenience  of  the  student,  but 
it  should  be  borne  in  mind  that  these  are  only  average 
values,  and  in  many  special  cases  must  be  modified.  The 
factor  of  common  sense  must  be  used  in  determining  unit- 
working  stresses. 

TABLE   III 
FACTORS  OF  SAFETY 


Material. 

Steady  Stress, 
Buildings,  etc. 

Varying  Stress, 
Bridges. 

Repeated  or 
Reversed 

Stresses, 
Machines. 

Steel,  Hard  

5 

8 

15 

Steel,  Structural  

4 

6 

10 

Wrought  Iron 

4 

6 

10 

Cast  Iron  

6 

10 

20 

Timber. 

8 

10 

15 

Brick  and  Stone 

15 

25 

30 

PROS.  16.  A  block  of  wood  10  ins.  square  standing  on  end 
supports  50,000  Ibs.  What  is  the  factor  of  safety? 

PROS.  17.  A  boiler  stay-bolt  f  in.  in  diameter  sustains  a  load 
of  6000  Ibs.  Find  its  factor  of  safety.  (Use  root  of  thread.) 

PROB.  18.  The  mild  steel  piston  rod  of  a  12x18  in.  engine  is 
2\  ins.  in  diameter.  The  maximum  steam  pressure  is  100  Ibs.  per 
sq.  in.  Find  the  factor  of  safety  when  the  rod  is  in  tension. 


ART.  8.    SHEAR 

A  tensile  load  produces  an  elongation  of  the  specimen  with 
a  resulting  reduction  in  area.  As  the  load  passes  the  elastic 
limit  the  elongation  increases  at  a  greater  rate,  with  the 
result  that  specimen  begins  to  "  neck  "  down,  that  is,  the 
area  toward  the  center  of  the  specimen  is  reducing  at  a 


16 


STRENGTH  OF  MATERIALS 


greater  rate  than  anywhere  else.     This  is  referred  to  as  local 
elongation,  and  is  clearly  shown  in  Fig.  7. 


FIG.  7. 

The  unit  working  stress  in  tension  is  found  by  dividing 
the  total  load  by  the  original  area  of  the  specimen.  A  com- 
pressive  load  causes  a  shortening  of  the 
specimen  with  a  tendency  to  cause  rupture 
by  a  splitting  or  shearing,  provided  the 
specimen  is  not  too  long,  in  which  case  a 
bending  action  will  result.  The  usual 
method  of  failure  in  pure  compression  is 
shown  in  Fig.  8.  Here  again  the  unit 
compressive  stress  equals  the  total  load 
divided  by  the  original  area. 

plG  8  A    shearing   load    tends   to   produce   a 

cutting  across.  Thus,  in  the  case  of  *a  rivet 
the  unit  shear  would  equal  the  load  divided  by  the  cross- 
section  of  the  rivet. 


TABLE  IV 

ULTIMATE  SHEARING  STRENGTH 


Material. 

Ultimate  Strength, 
Pounds  per 
Square  Inch. 

f  With  Grain.  . 

600 

Timber  <   . 
1  Across  Gram  

Structural  Steel  

3,000 
50,000 

Wrought  Iron  
Cast  Iron  
Stone    .    . 

40,000 
20,000 
1  500 

SIMPLE  STRESSES 


17 


Table  IV  gives  the  ultimate  strength  in  shear  of  various 
materials.  There  are  some  cases  where  the  calculation  of 
the  unit  shear  is  not  so  easy  as  in  the  case  of  tension  and 
compression.  Consider  the  case  of  a  'bolt  where  the  shank 

of  the  bolt  is  put  in  tension  (see  Fig.  9).     The  unit  tensile 

p 

stress  on  the  shank  of  the  bolt  =  —      — ^.     The  head  of  the 

.7854d2 

bolt  tends  to  tear  away  from  the  shank,  setting  up  a  shear 
between  the  head  of  the  bolt  and  the  shank.  Now  assume 
that  the  head  and  shank  are  two  separate  pieces  and  that  the 


FIG.  9. 

shank  slips  into  a  hole  in  the  head  of  the  bolt.  It  is  evident 
that  the  area  of  contact  between  the  two  will  be  the  circum- 
ference of  the  shank  times  the  length  of  the  head.  Hence 
the  shearing  area  in  this  .case  becomes  3.1416XdX/i. 

EXAMPLE.     Let  P  =  5000 ; 
d  =  l  in.; 


Find  the  unit  tensile  stress  on  the  shank  of  bolt  and  the 
unit  shearing  stress  on  the  head  of  the  bolt. 

Let     &  =  unit  tensile  stress; 
/Ss  =  unit  shearing  stress. 


18  STRENGTH  OF  MATERIALS 


rh       Q      P  5000         5000     11Qonl, 

Then  ^  =  X, =  .  7854  X 12  =  ^442  =  11>320  lbs>  Per  S(l-  m- 

„      P  .5000  5000 


PROB.  19.  A  rivet  1  in.  in  diameter  is  subjected  to  a  load  of 
10,000  Ibs.  If  the  rivet  is  in  single  shear  find  the  unit  shearing  stress. 

PROB.  20.  Assume  same  rivet  as  in  Prob.  19  to  be  in  double 
shear.  Find  the  unit  shearing  stress. 

PROB.  21.  A  wrought-iron  bolt  1|  ins.  in  diameter  has  a  head 
If  ins.  long.  Find  (a)  the  unit  stress  tending  to  shear  the  head  and 
(6)  the  unit  tensile  stress  on  the  shank  of  the  bolt,  when  subjected 
to  a  pull  of  15,000  Ibs. 

ART.  9.     SIMPLE  PROBLEMS  IN  DESIGN 

In  this  article  is  presented  a  series  of  problems  involving 
simple  direct  stresses.  The  student  is  urged  in  every  case 
to  follow  some  general  plan  of  attack  in  the  solution  of  these 
and  other  problems.  In  the  judgment  of  the  author  there 
are  four  essential  steps  in  the  solution  of  any  problem: 

First— Think. 

Get  a  clear  mental  picture  of  what  the  problem 
involves. 

Second — Analyze. 

Figure  out  exactly  what  it  is  that  you  are  trying  to 
find  and  then  determine  what  elementary  principle 
underlies  the  problem.  A  free-hand  sketch  showing 
the  details  of  the  problem  is  very  helpful  in  this  regard. 

Third— Equate. 

Set  down  whatever  equations  are  involved.  Do 
not  memorize  equations,  but  get  a  firm  grip  on  fun- 
damental principles.  An  equation  is  simply  an  alge- 
braic expression  of  a  fundamental  principle. 


SIMPLE  STRESSES, 


19 


Fourth— Solve. 

Apply  rules  of  mathematics  and  make  it  a  point  to 
solve  in  the  simplest  manner  possible.  Keep  the 
unknown  quantity  on  the  left-hand  side  of  the  equation, 
and  all  known  factors  on  the  right-hand  side. 

Get  the  habit  of  working  problems  with  the  idea  of 
filing  them  away  for  reference  in  the  future.  Specify 
clearly  what  every  item  in  your  solution  signifies. 
Cheek  your  results  and  then 
acquire  the  feeling  of  confidence 
that  your  answer  is  correct, 
and  do  not  rely  upon  the  other 
fellow's  result. 

PROB.  22.  In  Fig.  10  the  pull  on 
the  rod  AB  is  25,000  Ibs.  Unit  working 
stress  of  the  wood  in  compression  is  800 
Ibs.  per  sq.  in.  Find  diameter  of  the  cast- 
iron  washer  if  the  hole  in  the  washer  is 
2|  ins.  diameter. 

PROB.  23.    Fig.  11  shows  a  wood  test  specimen.    If  the  ulti- 
mate tensile  strength  of  wood  is  10,000  Ibs.  per  sq.  in,  and  the 


Cast 
Iron 


FlG.  10. 


T 


I 


FIG.  11. 


ultimate  shearing  strength  is  600  Ibs.  per  sq.  in.,  determine  the 
manner  in  which  the  specimen  will  fail. 

PROB.  24.  The  steam  pressure  in  a  boiler  is  125  Ibs.  per  sq.  in. 
Each  boiler  stay  supports  an  area  of  7|  Xof  ins.  Find  the  diameter 
of  the  stay  if  the  unit  stress  is  not  to  exceed  5800  Ibs.  per  sq.  in. 

PROB.  25.  A  wrought-iron  rod  1|  ins.  diameter  broke  under  a 
load  of  67,610  Ibs.  Find  the  ultimate  tensile  strength  of  the 
wrought  iron. 


20 


STRENGTH  OF  MATERIALS 


PROB.  26.  What  pull  will  be  required  to  break  a  structural  steel 
rod  2|  ins.  in  diameter? 

PROB.  27.  A  load  of  4000  Ibs.  is  suspended  as  shown  in  Fig.  12. 
Find  the  diameter  of  the  steel  rods  A  and  B,  using  a  factor  of  safety 
of  6. 


PROB.  28.  In  Fig.  6  assume  AB  to  be  a  square  wooden  strut 
and  consider  it  as  under  pure  compression.  Find  the  size  of  strut, 
using  a  factor  of  safety  of  10. 

PROB.  29.  If,  in  Fig.  9,  the  diameter  d  of  the  bolt  is,  1^  ins.  and 
the  length  of  head  his  1  in.,  determine  the  manner  in  which  the  bolt 
will  fail. 

PROB.  30.  A  cast-iron  test  specimen  1  in.  in  diameter  broke 
under  a  load  of  75,000  Ibs.  Find  the  ultimate  compressive  strength 
of  cast  iron. 


CHAPTER   III 
MATERIALS  OF  CONSTRUCTION 

ART.  10.     TESTING  MACHINES 

THE  physical  properties  of  the  various  materials  of  con- 
struction are  determined  experimentally  by  means  of  a  testing 
machine,  a  common  form  of  which  is  sho,wn  in  Fig.  13.  By 
means  of  this  machine  tests  in  tension,  compression,  shear, 
and  bending  can  be  made.  The  machine  consists  of  a  heavy 
platen  A,  mounted  upon  knife  edges  B,  which  in  turn  are 
secured  to  a  system  of  levers  connected  to  a  weighing  beam 
C,  upon  which  is  mounted  the  sliding  weight  D.  Power  is 
applied  by  gearing  to  the  four  vertical  screws  which  operate 
the  movable  head  E.  These  screws  pass  through  holes  in 
the  fixed  head,  but  are  independent  of  the  weighing  mechan- 
ism. Mounted  upon  the  platen  are  four  heavy  metal  struts 
Fj  which  support  the  fixed  head  H.  There  is  a  tapered  slot 
in  both  the  fixed  and  movable  heads.  In  these  slots  are 
inserted  tapered  jaws  for  holding  the  specimen.  The  ma-, 
chine  is  provided  with  several  speeds  to  facilitate  operation. 

The  up-and-down  motion  of  the  movable  head  is  con- 
trolled by  the  lever  L,  which  operates  a  clutch.  An  open 
belt  runs  on  the  pulley  M.  When  the  lever  L  is  thrown  to 
the  left  the  pulley  M  operates  the  drive  shaft,  and  when  L  is 
thrown  to  the  right  the  pulley  R,  on  which  a  crossed  belt 
operates,  connects  with  the  drive  shaft.  The  pulley  M 
causes  downward  motion  of  the  head  E,  and  the  pulley  R 
causes  upward  motion. 

21 


22 


STRENGTH  OF  MATERIALS 


When  a  slow  speed  is  desired  the  lever  L  is  put  in  neutral 
position  (directly  vertical);  the  hand-wheel  G  is  turned  to 
the  right,  which  causes  the  friction  gear  at  P  to  operate. 
This  results  in  a  very  slow  rotation  of  the  drive  shaft  and 
consequently  slow  downward  motion  of  the  movable  head. 


FIG.  13. 

The  lever  S  is  used  to  change  gears  when  a  quick  motion  of 
the  head  is  desired,  but  this  lever  must  never  be  operated 
when  the  machine  is  running. 

In  the  case  of  a  tension  test  the  movable  head  works 
downward,  the  specimen  being  secured  in  the  jaws  of  both 
the  fixed  and  the  movable  heads.  This  elongates  the  speci- 
men and  transmits  the  load  through  the  fixed  head  to  the 


MATERIALS  OF  CONSTRUCTION  23 

struts,  to  the  platen,  and  thence  to  the  weighing  mechanism. 
The  poise  D  is  operated  by  the  hand  wheel  N,  so  that  the  load 
on  the  specimen  can  be  determined  by  keeping  the  weighing 
beam  in  balance  at  all  times.  The  elongation  is  measured 
by  means  of  an  extensometer  which  is  attached  to  the  speci- 


FIG.  14. 


men  under  test.     A  good  form  of  this  instrument  is  chown 
in  Fig.  14. 

In  conducting  a  compression  test  a  flat  plate  is  placed 
under  the  movable  head  and  the  specimen  to  be  tested  is 
placed  between  this  plate  and  the  platen.  Power  is  applied 


24 


STRENGTH  OF  MATERIALS 


and  the  load  on  the  specimen  is  transmitted  directly  to  the 
platen  and  thence  to  the  weighing  mechanism.  A  counter 
weight  is  provided  to  balance  the  weight  of  the  levers,  platen, 
etc.  Before  starting  either  a  tension  or  compression  test  it  is 
essential  that  beam  is  floating  when  the  riding  weight  D 
is  on  the  zero  mark.  In  compression  tests  the  deformation 
is  recorded  by  means  of  a  compressometer,  which  is  con- 
structed somewhat  similarly  to  the  extensometer. 

The  capacity  of  a  machine  is  expressed  in  pounds,  for 
example,  a  250,000-lb.  machine  indicates  one  that  is  capable 
of  breaking  a  specimen  requiring  a  load  of  250,000  Ibs.  to 
produce  rupture.  These  machines  are  either  belt  or  motor 
driven.  In  some  cases  the  machine  is  provided  with  an 
attachment  for  automatically  recording  on  a  chart  the  load 
and  deformation  on  the  specimen.  For  more  detailed  expla- 
nation of  these  machines  the  student  is  referred  to  the  trade 
catalogues. 


ART.  11.     TENSION  AND  COMPRESSION  TESTS 

The  specimens  used  for  tension  tests  vary  in  size  and 
shape  with  the  material  to  be  tested  and  the  purpose  for 


St'd  Thread 


FIG.  15. 

which  it  is  intended.  Fig.  15  shows  a  form  used  for  structural 
steel  Fig.  15a  for  cast  iron  and  Fig.  16  a  form  used  for  testing 
boiler  plates. 

Fig.  17  gives  a  typical  curve,  showing  the  relation  between 
the  total  load  and  the  total  elongation  for  structural  steel. 


MATERIALS  OF  CONSTRUCTION 


25 


The  five  significant  results  of  a  tension  test  are: 

(1)  The  ultimate  tensile  strength. 

(2)  The  elastic  limit. 

(3)  The  per  cent  elongation. 

(4)  The  per  cent  reduction  of  area. 

(5)  The  modulus  of  elasticity. 


U.S.St'd 


K..R. 


U.S.St'd  ^ 


FIG.  15a. 


The  ultimate  strength  is  found  by  dividing  the  maximum 
load  in  pounds  required  to  rupture  the  specimen  by  the  orig- 
inal area  of  the  specimen  in  square  inches. 


-18- 


-12- 


Al 


-H* 


FIG.  16. 

The  elastic  limit  is  found  by  dividing  the  load  in  pounds, 
at  the  elastic  limit,  by  the  original  area  of  the  specimen  in 
square  inches. 

The  per  cent  elongation  is  determined  from  the  ratio  of  the 
total  elongation,  which  is  the  difference  between  the  final  and 
the  original  lengths  of  the  part  under  test,  to  the  original 
length. 

The  per  cent  reduction  in  area  is  the  difference  between  the 
original  and  final  areas  divided  by  the  original  area.  In 


26 


STRENGTH  OF  MATERIALS 


both  the  case  of  elongation  and  reduction  of  area  the  ratio 
must  be  multiplied  by  the  factor  100  to  express  the  result  in 
per  cent. 

The  modulus  of  elasticity  is  the  ratio  of  the  unit  stress  at 
the  elastic  limit  to  the  unit  deformation  at  the  elastic  limit. 
In  the  case  of  a  tension  test  the  unit  deformation  equals  the 
total  elongation  at  the  elastic  limit  divided  by  the  original 
length  of  the  part  under  test.  If  the  specimen  breaks  near 
either  end  a  correction  must  be  made  for  the  local  elongation. 


1.2  1.6  2.0 

Elongation— Inches 

FIG.  17. 


2.4 


Compression  tests  are  conducted  in  the  same  manner  as 
tension  tests,  and  the  results  are  figured  on  the  same  basis. 
Fig.  18  gives  a  typical  curve  for  a  compression  test  of  cast 
iron. 

EXAMPLE.  Find  the  five  significant  results  of  the  ten- 
sion test  shown  in  Fig.  17,  where  the  original  diameter  of 
test  specimen  was  .78  in.  and  the  part  under  test  8  ins.  in 
length.  The  final  diameter  was  .553  in.  and  the  final  length 
10.232  ins. 

SOLUTION.  From  the  curve  the  maximum  load  is  found 
to  be  29,000  Ibs.  and  the  load  at  the  elastic  limit  19,000  Ibs. 
The  total  elongation  at  the  elastic  limit  is  0.0104  in. 


MATERIALS  OF  CONSTRUCTION 


27 


ijjina!  Diam. 
ngth 


0  0.1  0.2  0.3 

Total  Deformation  -  Inches 


FlG.  18. 


Hence: 


9QOOO 
(1)  The  ultimate  strength  =  -        =  60,660  Ibs.  per  sq.  in. 


(2)  The  elastic  limit  -          -  =  40,000  Ibs.  per  sq.  in. 


10  939  _  8 

(3)  Per  cent  elongation  =  -  -  X  100  =  28. 

o 

(4)  Per  cent  reduction  of  area  =  —          -  X  100  =  50. 

.478 

40000 

(5)  Modulus  of  elasticity  =  -^^  =  30,000,000. 


PROB.  31.  A  2-in.  cube  of  yellow  pine  broke  under  a  com- 
pressive  load  of  28,000  Ibs.  The  load  at  elastic  limit  was  12,000  Ibs. 
The  deformation  at  the  elastic  limit  was  .004  in.  Find  the  ultimate 
compressive  strength,  the  elastic  limit,  and  the  modulus  of  elasticity. 

PROB.  32.  A  test  specimen  of  boiler  plate  2x|  in.  in  section 
broke  under  a  load  of  52,000  Ibs.  The  elongation  in  8  ins.  was  2.2 
ins.  Find  the  tensile  strength  and  the  per  cent  elongation. 


28  STRENGTH  OF  MATERIALS 


ART.  12.    CAST  IRON 

Iron,  which  constitutes  the  most  important  element  in 
modern  industry,  never  occurs  free  in  nature,  but  is  always 
combined  with  other  elements  and  impurities.  The  primary 
source  of  iron  is  from  the  enormous  deposits  of  iron  ore  in 
this  and  other  countries.  This  iron  ore  is  placed,  together 
with  a  fuel,  usually  coke,  and  a  flux,  usually  limestone,  in  a 
large  vertical,  conical-shaped  retort  called  the  blast  furnace. 
Under  the  action  of  a  blast  of  hot  air  the  charge  in  the  fur- 
nace is  gradually  reduced  to  a  molten  state,  and  the  iron, 
together  with  other  elements,  collects  in  the  bottom  or  hearth 
of  the  furnace.  At  stated  intervals  the  furnace  is  tapped 
and  the  iron  conducted  either  into  large  ladles,  or  run  directly 
into  molds,  producing  "  pig  iron."  At  a  point  somewhat 
above  the  metal  tap  is  another  tap  for  the  removal  of  the 
"  slag."  This  slag  is  a  result  of  the  combination  of  the  im- 
purities in  the  iron  ore  with  the  "  flux,"  and  is  produced  in 
such  large  quantities  that  its  final  disposition  is  no  small 
problem  in  the  modern  iron  industry. 

If  the  iron  is  not  to  be  used  directly  for  the  manufacture 
of  steel  it  is  run  into  molds  made  of  sand.  These  molds  pro- 
duce a  pig  weighing  about  80  Ibs.  This  pig  iron  is  shipped 
from  the  blast  furnace  to  the  various  foundries,  where  it  is 
placed,  together  with  a  fuel  and  flux,  into  a  "  cupola."  Here 
under  the  action  of  a  blast  of  air  the  pig  iron  is  reduced 
to  a  molten  state  and  the  impurities  again  partially  removed, 
From  the  cupola  the  iron  is  poured  into  ladles  and  thence 
into  whatever  forms  of  molds  that  may  have  been  prepared. 
The  student  must  keep  in  mind  the  fact  that  "  pig  iron  "  is 
the  product  of  the  blast  furnace  and  cast  iron  is  the  product 
of  the  cupola  in  the  iron  foundry. 

The  grade  of  cast  iron  can  be  closely  regulated  by  the 
cupola  charge.  Frequently  pig  iron  is  charged,  together 


MATERIALS  OF  CONSTRUCTION  29 

with  better  grades  of  cast  iron,  into  the  cupola.  In  this  way 
the  final  composition  of  the  cast  iron  can  be  regulated. 

Cast  iron  consists  of  metallic  iron  with  at  least  1.5  per 
cent  carbon.  It  also  contains  silicon,  sulphur,  phosphorus, 
manganese,  and  other  elements.  Usually  there  will  be  about 
90  to  92  per  cent  of  iron  and  10  to  8  per  cent  of  other  elements 
in  cast  iron.  The  carbon  occurs  either  in  the  free  or  "  gra- 
phitic "  form  or  in  the  "  combined  "  state.  The  per  cent  of 
carbon  materially  effects  the  physical  properties  of  the  iron; 
if  the  carbon  is  in  the  graphite  or  free  state,  a  gray  iron  will 
be  the  result;  if  the  carbon  is  in  the  combined  state,  a  white 
iron  will  result. 

Silicon  in  cast  iron  prevents  the  combining  of  the  carbon, 
and  a  gray  iron  results.  The  per  cent  of  silicon  varies  from 
.5  to  4  per  cent.  The  silicon  tends  to  prevent  blowholes  in 
the  castings,  thus  producing  an  iron  of  greater  density. 
Sulphur  in  cast  iron  should  never  exceed  .15  per  cent;  it  is 
considered  an  undesirable  element. 

Phosphorus  tends  to  make  cast  iron  more  fluid,  but  also 
induces  great  brittleness.  For  good  strong  castings  phos- 
phorus should  not  exceed  .6  per  cent.  Manganese  tends  to 
produce  hardness  in  the  iron. 

Cast  iron  finds  its  greatest  use  in  machine  frames,  bed 
plates,  etc.,  on  account  of  its  high  compressive  strength  and 
the  ease  with  which  the  iron  can  be  cast  into  almost  any  size 
or  shape.  Cast  iron  is  not  used  in  tension,  owing  to  its  low 
tensile  strength,  and  to  the  facts  that  its  per  cent  elongation 
is  negligible ;  it  has  no  well-defined  elastic  limit  in  tension. 

PROB.  33.  A  cast-iron  specimen  1  in.  in  diameter  and  2  in. 
long  broke  under  a  load  of  18,000  Ibs.  Find  the  ultimate  tensile 
strength  of  cast  iron. 

PROB.  34.  A  cast-iron  bearing  plate  supports  a  load  of  50,000 
Ibs.  Find  its  cross-sectional  area  so  that  the  unit  compressive 
stress  shall  not  exceed  one-quarter  the  elastic  limit. 


30  STRENGTH  OF  MATERIALS 


ART.  13.    WROUGHT  IRON 

Wrought  iron  is  defined  by  Campbell  as  slag-bearing, 
malleable  iron  which  does  not  harden  materially  when  sud- 
denly cooled. 

Wrought  iron  is  manufactured  by  heating  pig  iron, 
together  with  iron  ore,  scrap  iron,  and  other  fettling  material 
in  a  puddling  furnace  which  is  lined  or  "  fettled  "  with  iron 
oxides  in  the  form  of  roll  scale,  high-grade  iron  ore,  or  roasted 
puddle  cinder.  The  furnace  is  of  the  reverberatory  type, 
with  a  capacity  of  from  300  to  1500  Ibs.  The  fuel  used  is 
either  gas  or  bituminous  coal,  and  the  flame  is  allowed  to 
play  on  the  "  bath  "  of  metal  in  the  furnace  hearth. 

The  pig  iron  used  for  manufacture  of  wrought  iron  is 
known  as  forge  iron,  and  contains  about  1  per  cent  of  silicon, 
0.5  per  cent  manganese,  less  than  1  per  cent  of  phosphorus, 
and  not  over  0.1  per  cent  of  sulphur,  The  larger  the  per  cent 
of  impurities  in  the  charge  the  greater  the  loss  of  metal  during 
the  process. 

The  melting-down  stage  requires  from  thirty  to  thirty-five 
minutes,  after  which  there  is  a  stage  of  from  seven  to  ten 
minutes  during  which  very  high  grade  iron  ore  is  added, 
during  which  time  the  damper  is  put  on  so  as  to  cool  the 
charge  and  oxidize  the  impurities  before  the  carbon.  This 
produces  the  "  boil,"  which  lasts  from  twenty  to  twenty-five 
minutes,  during  which  time  the  puddler  stirs  the  charge  with 
a  long  iron  bar,  the  slag  pouring  out  into  a  slag  bucket. 
During  the  next  fifteen  to  twenty  minutes  the  "  balling  " 
period  occurs,  at  which  time  the  puddler  divides  the  bath 
into  a  number  of  balls  which  are  removed  from  the  furnace 
by  means  of  a  pair  of  tongs  suspended  from  an  overhead 
track.  These  balls  are  taken  to  the  "  squeezer,"  where  part 
of  the  slag  is  removed  by  passing  the  balls  through  an  eccen- 
tric vertical  roll,  leaving  the  metal  in  the  form  of  a  rectangular 


MATERIALS  OF  CONSTRUCTION 


31 


bar.  This  bar  is  then  put  through  the  rolls,  producing  what  is 
called  "  muck  "  bar,  which  contains  a  large  percentage  of 
slag  and  impurities. 

The  muck  bars  are  placed  in  a  reheating  furnace,  from 
which  they  are  taken  and  again  passed  through  rolls,  produc- 
ing merchant  bar,  or  commercial  wrought  iron.  When  mer- 
chant bar  is  sheared,  piled,  and  rerolled  the  resulting  mate- 
rial is  called  double  refined  iron.  Coal  is  generally  used  as 
fuel  in  the  puddling  furnace.  When  charcoal  is  used  as  fuel 
and  the  air  forced  in  through  tuyeres  a  better  product  is  the 
result.  This  material  is  commercially  known  as  charcoal 
iron. 

Wrought  iron,  owing  to  its  method  of  manufacture,  gives 
a  fibrous  fracture.  It  has  a  fairly  high  tensile  strength  with 
somewhat  less  per  cent  elongation  than  steel.  Wrought  iron 
is  easily  welded,  and  for  this  reason  finds  its  greatest  use 
in  parts  of  machines 
where  shaping  by  forg- 
ing is  necessary. 

The  product  of  the 
rolling  mill  is  bar  iron 
and  plate  iron.  Owing 
to  the  method  of 
manufacture  the  size 
of  plate  is  limited. 
The  thickness  depends 
upon  the  type  of 
service  for  which  the 
iron  is  intended. 

PROB.  35.  Fig.  19 
shows  a  simple  band  JPIG<  19. 

brake.      Maximum  ten- 
sion  on   the  wrought-iron  strap  is  2000  Ibs.      If  the   thickness 
of  the  strap  is  TQ  in.,  find  the  breadth  using  a  factor  of  safety 
of  10. 


32  STRENGTH  OF  MATERIALS 

PROB.  36.    Fig.  20  shows  a  form  of  malleable-iron  chain.    The 
links  are  2  ins.  wide  and  |  in.  thick.     Find  the  safe  load  that  the 


FIG.  20. 

chain  can  carry  so  that  the  unit  tensile  stress  shall  not  exceed  6000 
Ibs.  per  sq.  in. 

ART.  14.     STEEL 

By  the  term  steel  is  meant  the  product  of  the  cementation 
process  or  the  malleable  compounds  of  iron  made  in  the  cru- 
cible, the  converter,  or  the  open-hearth  furnace. 

Crucible  Steel.  Wrought  iron  and  steel  scraps,  together 
with  charcoal  or  pig  iron,  and  other  ingredients,  dependent 
upon  the  product  desired,  are  placed  in  a  covered* -crucible 
made  of  about  50  per  cent  graphite  and  50  per  cent  clay. 
These  crucibles  hold  between  80  and  100  Ibs.  of  metal, 
and  last  from  six  to  eight  heats.  The  crucibles  are  placed 
in  rows  in  a  regenerative  furnace  heated  by  producer  gas  or 
coke. 

The  process  requires  from  four  to  five  hours,  at  the  end  of 
which  time  the  crucibles  are  withdrawn  by  a  "  puller  out," 
who  straddles  the  pit  in  which  the  crucibles  are  placed  and 
draws  them  out  with  a  pair  of  tongs.  The  crucibles  are  then 
emptied  into  a  large  ladle  from  which  the  metal  is  poured  into 
ingots  of  any  desired  shape. 

Crucible  steel  is  largely  used  for  the  making  of  instru- 
ments and  tools  for  use  in  high-speed  machine  work.  There 
are  about  seventy-five  different  alloy  steels  made  by  the 
process,  the  alloys  being  nickel,  chromium,  vanadium,  etc. 


MATERIALS  OF  CONSTRUCTION  33 

Bessemer  Process.  The  Bessemer  converter  consists  of  a 
cylindrical  cast-iron  vessel  lined  with  firebrick  and  having  a 
conical-shaped  mouth-piece.  The  bottom  consists  of  a 
casting  provided  with  a  series  of  tuyeres  or  air  ports  through 
which  a  blast  of  air  is  admitted. 

The  pig  iron  is  brought  from  the  blast  furnace  in  a  molten 
state  and  placed  in  a  large  mixer,  holding  200  to  300  tons, 
heated  by  oil  or  gas.  The  converter  is  placed  in  a  horizontal 
position  and  a  charge  of  molten  metal  admitted.  The  blast 
is  turned  on  and  the  converter  put  in  a  vertical  position. 
The  air  blowing  up  through  the  molten  iron  burns  out  the 
carbon  and  other  impurities.  This  requires  from  ten  to 
twenty  minutes.  During  this  period  considerable  of  the  iron 
is  blown  away.  The  quality  of  the  product  is  determined 
by  observation  of  the  flame  emerging  from  the  converter. 
After  all  impurities  have  been  burned  out  a  small  amount  of 
"  spiegeleisen  "  is  admitted  in  order  to  bring  up  the  carbon 
to  the  desired  content.  At  the  end  of  the  "  blow  "  the  con- 
verter is  placed  in  a  horizontal  position,  the  blast  shut  off, 
and  the  metal  poured  into  a  large  ladle,  from  which  it  is 
poured  into  ingots.  The  production  of  steel  by  the  Bessemer 
process  is  being  replaced  by  the  open-hearth  method. 

Open-hearth  Process.  Fig.  21  shows  a  typical  form  of  fur- 
nace used  in  the  manufacture  of  steel.  This  consists  of  an 
open  hearth  K,  lined  with  either  dolomite  (limestone)  or 
silica  (sand),  depending  upon  the  per  cent  of  phosphorus  in 
the  material  constituting  the  charge.  Producer  gas  is  used 
as  the  fuel,  and  is  preheated  by  passing  through  the  "  regen- 
erative "  chamber  F.  The  air  passes  through  the  chamber 
E,  where  it  is  also  preheated.  The  air  and  gas  combine  in 
chamber  H,  producing  a  very  high  temperature  over  the 
hearth.  The  depressed  roof  causes  the  flame  to  impinge  on 
the  charge.  The  products  of  combustion  pass  out  through 
the  chamber  I  to  the  chambers  E  and  F,  which  are  filled  with 
layers  of  firebrick.  These  chambers  become  very  hot.  At 


34 


STRENGTH  OF  MATERIALS 


the  end  of  a  given  interval  of  time,  usually  twenty  to  thirty 
minutes,  a  set  of  valves  are  operated  which  cause  the  air  and 
gas  to  come  in  through  the  right-hand  chambers  and  pass 
out  through  the  left-hand  chambers.  In  this  way  the  waste 
heat  in  the  products  of  combustion  is  utilized  in  preheating 
the  air  and  gas. 

The  charge  consists  of  all  kinds  of  steel  and  iron  scrap 
together  with  a  certain  amount  of  flux,  depending  upon  the 


character  of  the  scrap.  In  other  cases  the  pig  iron  is  taken 
directly  from  the  blast  furnace  and  run  into  the  mouth  of 
the  open-hearth  furnace.  This  process  takes  from  five  to  six 
hours.  The  hearth  holds  from  50  to  100  tons  of  metal.  The 
per  cent  of  carbon  can  be  definitely  fixed  in  this  process,  as 
samples  can  be  drawn  off  from  time  to  time  and  tested.  At 
the  end  of  the  heat  the  steel  is  drawn  off  in  a  large  ladle  from 
which  it  is  poured  into  large  ingots,  which  are  later  put 


MATERIALS  OF  CONSTRUCTION 


35 


through  the  rolls  and  formed  into  the  various  structural 
shapes  such  as  rails,  I  beams,  angles,  etc.  The  open-hearth 
process  gives  a  steel  of  more  uniform  composition,  and  for 
this  reason  is  rapidly  replacing  the  Bessemer  process. 

The  physical  properties  of  steel  depend  upon  the  method 
of  manufacture  and  upon  the  chemical  composition  of  the 
steel.  The  carbon  content  has  much  to  do  with  the  hardness 
of  the  steel.  Silicon  increases  the  strength  and  reduces  the 
ductility  of  steel.  Manganese  increases  the  hardness  and 
raises  the  elastic  limit  and  ultimate  strength.  Phosphorus 
increases  the  static  strength,  but  reduces  the  resistance  to 
shock.  Table  V  shows  the  composition  of  steel  plates  used 
for  various  purposes: 

TABLE  V 


Quality. 

Carbon. 

Manganese. 

Sulphur. 

Phosphorus. 

Firebox  
Boiler  

.16 

.18 

.35  to    .50 
.35  to    .60 

Not  over  .   04 
Not  over  .045 

Not  over  .  02 
Not  over  .  04 

Flange 

18 

35  to    .60 

Not  over  .045 

Not  over  .  04 

Ship  
Tank  

.15 
.10 

.35  to    .65 
.40 

Not  over  .060 
Not  over  .100 

Not  over  .08 
Not  over  .  12 

PROB.  37.    A  steel  bar  1|  ins.  in  diameter  is  subjected  to  a  ten- 
sile pull  of  18,000.    Find  the  unit  tensile  stress  and  the  factor  of 

safety. 


ART.  15.    TIMBER,  STONE,  AND  BRICK 

Timber  varies  more  widely  in  general  characteristics  than 
any  of  the  common  building  materials.  The  structure  of 
wood  affords  the  only  positive  means  of  identifying  the  dif- 
ferent kinds.  Color,  weight,  and  smell  are  all  functions  of 
the  structure  of  the  wood.  In  the  lumber  trade  timber  is 
divided  into  two  classes,  "  hard  woods  "  and  "  soft  woods." 
The  terms  "  fine  grained,"  "  coarse  grained,"  "  straight 


36 


STRENGTH  OF  MATERIALS 


grained,"  and  "  cross  grained,"  refer  to  the  nature  of  the 
annual  rings,  whether  they  are  narrow  or  wide,  and  to  the 
fibers,  whether  they  are  parallel  or  twisted  relative  to  the  axis 
of  the  limb  or  tree. 

Timber  when  cut  contains  a  high  percentage  of  moisture, 
which  is  removed  by  drying  the  lumber  in  kilns  which  are 
heated  by  steam.  Hard  woods  such  as  oak,  ash,  maple,  etc., 
are  air-seasoned  from  three  to  six  months  to  allow  a  gradual 
shrinkage  before  being  finally  placed  in  the  kiln.  Seasoning 
increases  the  strength  of  timber,  but  once  seasoned,  timber 
should  not  be  exposed  to  the  weather.  Knots  reduce  the 
strength  of  timber.  If  possible  the  knotty  side  should  be 
in  compression.  Table  VI  gives  the  strength  of  the  more 
common  woods.  These  values  are  average  only  and  may 
vary,  depending  upon  the  character  of  the  timber.  Gen- 
erally the  denser,  close-grained  woods  are  the  stronger.  A 
study  of  the  fracture  of  the  various  woods  when  subjected 
to  compression  will  help  the  student  to  classify  timber  by  its 
structure.  The  elastic  limit  is  not  well  defined,  especially  in 
transverse  tests.  For  a  complete  discussion  of  tirhber  the 
student  is  referred  to  Johnson's  "  Materials  of  Construc- 
tion." 

TABLE  VI 


Weight,  Lbs. 

ULTIMATE 

STRENGTH. 

Shear 

per  Cu.  Ft. 

Tensile. 

Compressive. 

Grain. 

Yellow  Pine  (Georgia)  . 

38 

12,000 

7,000 

5,000 

White  Pine  

24 

7,000 

5,500 

2,000 

White  Oak  

50 

12,000 

7,000 

4,000 

Maple      

43 

18,000 

7,500 

Cypress 

28  7 

6000 

5000 

Ash 

40 

15000 

8000 

Chestnut  

41.2 

8,500 

4,000 

2,000 

Hemlock  

25 

6,000 

7,000 

2,500 

MATERIALS  OF  CONSTRUCTION  37 

Stone  used  for  structural  purposes  should  possess  the 
qualities  of  cheapness,  durability,  strength,  and  beauty. 
As  a  general  rule  the  densest,  hardest,  and  most  uniform 
stone  will  meet  these  requirements.  The  stone  on  fracture 
should  be  bright,  clean,  and  sharp  without  loose  grain  and 
free  from  any  dull,  earthy  appearance. 

The  crushing  strength  of  stone  is  found  by  placing  a  prism 
or  cube  of  the  material  in  a  testing  machine.  The  specimen 
will  generally  fail  by  shear  on  a  well-defined  angle.  This  is 
typical  of  all  brittle  materials.  Geological  classification 
divides  rocks  into  three  groups — igneous,  metamorphic,  and 
sedimentary.  Greenstone,  trap,  and  lava  are  examples  of 
igneous  rocks;  granite,  marble,  and  slate,  of  metamorphic; 
sandstone  and  limestone  of  sedimentary. 

Trap  is  the  strongest  of  building  stones,  but  is  little  used 
owing  to  its  toughness  and  difficulty  of  quarrying.  It  is  very 
durable. 

Granite  is  the  strongest  and  most  durable  of  all  the  stones 
commonly  used.  It  breaks  with  regularity  and  is  easily 
quarried  into  simple  shapes.  Its  extreme  hardness  and 
toughness  makes  it  an  expensive  stone  where  special  form  is 
desired,  hence  it  is  most  commonly  used  for  foundations, 
piers,  or  for  large  buildings.  The  largest  blocks  of  stone 
ever  quarried  have  been  of  granite,  one  case  being  noted 
where  a  shaft  115  ft.  long  and  10  ft.  square  at  the  base  and 
weighing  850  tons  was  quarried. 

Limestone  is  composed  chiefly  of  carbonate  of  lime. 
Limestones  vary  in  color  and  composition,  depending  upon 
the  character  of  the  deposit. 

Marble;  any  limestone  which  will  take  on  a  polish  is  called 
a  marble.  This  makes  a  most  beautiful  building  material 
and  is  used  for  interior  decoration.  Marble  is  extensively 
used  in  modern  plumbing  fixtures. 

Sandstones  are  so  called  because  they  are  made  up 
chiefly  of  sand,  cemented  and  consolidated.  Sandstones 


38 


STRENGTH  OF  MATERIALS 


vary  in  color  and  texture,  depending  upon  ingredients  such 
as  lime,  iron,  manganese,  and  alumina.  The  durability  of  the 
sandstones  varies  with  their  physical  and  chemical  properties. 
Sandstones  are  as  resistant  to  weather  as  granite  and  are 
less  affected  by  fire.  They  are  easily  wrought  into  shape 
with  hammer  and  chisel  and,  therefore,  constitute  the  most 
common  stone  used  for  building  purposes. 

The  weights  and  strengths  of  the  various  stones  are  given 
in  Table  VII: 

TABLE  VII 


Weight,  Lbs. 

ULTIMATE 

STRENGTH. 

per  Cu.  Ft. 

Compression. 

Shear. 

Sandstone  

150 

5,000 

1,400 

Limestone                     

160 

8,000 

1,700 

Granite 

165 

14,000 

2,500 

Trap.  . 

175 

16,000 

The  above  values  are  the  results  of  tests  on  small  speci- 
mens. A  test  of  an  actual  structure  would  show  values  of 
approximately  75  per  cent  of  the  above. 

Brick  is  made  of  common  clay,  consisting  mainly  of  silicate 
of  alumina,  by  submitting  it  to  a  temperature  which  con- 
verts it  into  a  semi-vitrified  mass.  Owing  to  its  ease  of  man- 
ufacture, transportation,  and  handling,  brick  is  used  in  place 
of  stone.  When  properly  made  it  is  almost  as  strong  as 
stone.  Professor  Baker  gives  the  following  six  requisites 
for  good  brick: 

1.  A  good  brick  should  have  plane  faces ;  parallel  sides, 

sharp  edges  and  angles. 

2.  It  should  be   of  fine,   compact,   uniform  texture; 

should  be  quite  hard,  and  should  give  a  clear 
ringing  sound  when  struck  a  sharp  blow. 


MATERIALS  OF  CONSTRUCTION  39 

3.  It  should  not  absorb  more  than  one-tenth  its  weight 

of  water. 

4.  Its  specific  gravity  should  be  two  or  over. 

5.  The  crushing  strength  of  half  brick  when  ground 

flat  and  pressed  between  thick  metal  plates  should 
be  at  least  7000  Ibs.  per  sq.  in. 

6.  Its  modulus  of  rupture  should  be  at  least  1000  Ibs. 

per  sq.  in. 

The  common  size  of  a  brick  in  this  country  is  8JX4X2J, 
thus  making  twenty-three  to  the  cubic  foot.  These  average 
about  4J  Ibs.  each.  Brick  is  sold  and  laid  by  the  thousand. 

When  used  as  paving  materials  both  brick  and  stone  must 
be  carefully  tested  for  their  abrasive  properties. 

PROB.  38.  Find  the  weight  of  a  stone  pier  whose  lower  base  is 
12x10;  upper  base  8x10  and  altitude  6  ft. 

PROB.  39.  Find  the  weight  of  a  13-in.  brick  wall  built  in  the 
shape  of  a  triangle,  whose  base  is  30  ft.  and  altitude  12  ft. 

PROB.  40.  Find  the  unit  stress  at  the  base  of  a  granite  shaft 
100  ft.  in  height  What  is  the  factor  of  safety? 

ART.  16.     PROPERTIES  OF  MATERTALS 

Wrought  iron  weighs  480  Ibs.  per  cu.  ft.  This  fact  gives 
rise  to  the  following  rules  for  determining  the  weights  of  bars 
or  other  parts  of  uniform  cross-section:  A  bar  of  wrought 
iron  1  yd.  long  and  1  sq.  in.  in  cross-section  weighs  10  Ibs. 
Steel  is  2  per  cent  heavier  and  weighs  10.2  Ibs.  per  yard. 
Cast  iron,  being  6  per  cent  lighter,  weighs  9.4  Ibs.  per  yd. 
For  example,  a  steel  I  beam  having  a  cross-section  of  5.88 
sq.  ins.  weighs  5.88  X  10.2  =  60  Ibs.  per  yd.  A  cast-iron  grate 
bar  6  sq.  ins.  in  cross-section  weighs  6X9.4  =  57.4  Ibs  per 
yard.  A  wrought-iron  boiler  plate  f  in.  thick  weighs  J  X 1 X 
10  ins.  =  7.5  Ibs.  per  yard  per  inch  width  of  plate.  Table 
VIII  gives  the  average  weights  in  pounds  per  cubic  foot  of 
the  various  materials  used  in  construction. 


40 


STRENGTH  OF  MATERIALS 


TABLE    VIII 


Material. 

Weight,  Lbs. 
per  Cu.  Ft. 

Specific 
Gravity. 

Steel.          

490 

7.05 

Wrought  iron  
Cast  iron                                                     .... 

480 
450 

7.69 

7  21 

Stone  
Brick,  pressed.  
Brick  common          •                                     .  . 

160 
140 
125 

2.60 

Brass 

500 

8  1 

Lead                              

710 

11  38 

Stone  masonry                                          .      .  . 

150 

Timber 

40 

64 

PROB.  41.  A  cast-iron  water  pipe  is  24  ins.  in  diameter.  The 
thickness  of  the  metal  is  1.5  in.  Find  the  weight  of  the  pipe  per 
linear  yard. 

PROB.  42.  The  bed  plate  of  planer  weighs  450  Ibs.  per  linear 
yard.  Find  its  cross-sectional  area. 

ART  17.     REVIEW  PROBLEMS 

PROB.  43.  Find  the  weight  of  a  yellow  pine  beam  8X12  ins. 
cross-section  and  20  ft.  long. 

PROB.  44.  A  steel  wire  1.3  in.  in  diameter  broke  under  a 
tensile  load  of  294,000  Ibs.  Find  its  ultimate  tensile  strength. 

PROB.  45.  A  cast-iron  bar  1  in.  in  diameter  broke  under  a  com- 
pressive  load  of  74,000  Ibs.  Find  the  ultimate  compressive  strength.  • 

PROB.  46.  A  block  of  maple  2x2x2  ins.  broke  under  a 
compressive  load  of  24,000  Ibs.  Find  the  ultimate  strength. 

PROB.  47.  A  boiler  plate  specimen  ^X2  ins.  broke  under  a 
load  of  55,000  Ibs.  The  elongation  in  8  ins.  was  2.5  ins.  Find  the 
ultimate  strength  and  the  per  cent  elongation. 

PROB.  48.  A  stone  pier  is  8  X 10  ft.  in  cross-section.  What  safe 
load  can  the  pier  carry  if  the  safe  bearing  power  of  the  soil  on  which 
the  pier  rests  is  2.5  tons  per  sq.  ft. 

PROB.  49.  In  a  tension  test  on  a  steel  specimen  0.8  in.  in  diam- 
eter the  total  elongation  was  found  to  be  .00264  in.  under  a  load  of 


MATERIALS  OF  CONSTRUCTION  41 

5000  Ibs.  Find  the  modulus  of  elasticity,  if  the  part  under  test 
was  8  ins.  long. 

PROB.  50.  A  hollow  steel  crank  shaft  is  10  ins.  outside  diameter 
and  7  ins.  inside  diameter.  Find  its  weight  if  the  shaft  is  12  ft.  long. 

PROB.  50cL  The  diameter  of  a  steel  bar  was  0.8  in.  before  test- 
ing and  0.56  in.  after  testing.  Find  the  reduction  in  area. 

PROB.  506.  Find  the  safe  load  that  can  be  carried  by  a  brick 
pier  24  X60  ins.,  using  a  factor  of  safety  of  12. 


CHAPTER  IV 
THEORY  OF  BEAMS 

ART.  18.    BEAM  REACTIONS 

BEAMS  are  divided  into  two  general  classes — simple  and 
cantilever,  depending  upon  their  method  of  support.  A 
beam  supported  at  both  ends  is  called  a  simple  beam,  while  a 
cantilever  beam  is  supported  at  one  end  only.  The  force 
exerted  by  the  part  supporting  the  beam  is  called  the  reaction 
of  the  support,  or  more  commonly  the  beam  reaction.  These 
reactions  are  determined  by  applying  the  principle  of  mo- 
ments discussed  in  Art  2,  Chapter  I. 


12  0 


1800 


-12- 


FIG.  22. 

AB,  Fig.  22,  represents  a  simple  beam  supported  at  A  and 
B,  loaded  as  shown.  Let  R\  equal  the  reaction  of  the  left- 
hand  support,  and  R2  the  reaction  of  the  right-hand  support. 
The  value  of  Ri  is  found  by  taking  moments  about  the  point 
R2,  thus: 

42 


THEORY  OF  BEAMS  43 

-6X1800+8X1200+12^1=0; 


#1  =  1700. 
In  like  manner  #2  is  found  by  taking  moments  about  Ri,  thus: 

4X1200+6X1800-12^2  =  0; 

12^2  =  15600; 
#2  =  1300. 

The  accuracy  of  these  results  can  be  checked  from  the  condi- 
tion of  equilibrium  that  the  algebraic  sum  of  the  vertical 
forces  equal  zero,  thus: 


-  1200-1800+#2  =  0; 
1700-1200-1800+1300  =  0; 
0  =  0. 

In  all  problems  dealing  with  the  reactions  of  simple  beams 
proceed  in  the  following  manner  :  First,  take  moments  about 
the  left-hand  support  to  determine  the  right-hand  reaction; 
second,  take  moments  about  the  right-hand  support  to  deter- 


-10- 


-25' 


FIG.  23. 

mine  the  left-hand  reaction;  third,  check  results  by  equating 
the  algebraic  sum  of  the  vertical  forces  on  the  beam  equal  to 
zero.  In  most  cases  the  weight  of  the  beam  must  be  included 
in  figuring  the  reaction.  A  beam  weighing  60  Ibs.  per  lin.  ft. 
is  loaded  as  shown  in  Fig.  23.  To  find  the  reactions  in  this 


44 


STRENGTH  OF  MATERIALS 


case  assume  the  weight  of  the  beam  equal  to  a  force  of 
25X60=1500  Ibs.  concentrated  at  the  center  of  the  beam. 
Taking  moments  about  Ri  there  results, 

3000  X5+1500X12J  +2000  X  15  -25#2=0; 

#2  =  2550. 
Taking  moments  about  #2  gives 

-  2000  X  10-  1500  X12J-  3000  X  20  +25#i=0; 


Using  SP  =  0  as  a  check,  there  results, 

#1  -3000-  1500  -2000+#2  =  0, 
or  3950-6500+2550  =  0; 

6500-6500  =  0. 

When  a  beam  overhangs  the  one  support  the  reactions 
can  be  found  just  the  same  as  in  the  previous  example.  Con- 
sider a  beam  weighing  40  Ibs.  per  lin.  ft.,  loaded  as  shown  in 
Fig.  24.  Assume  the  total  weight  =  40X30  =  1200  Ibs.  con- 


uu                               w 

(        \ 

W                                                                 3U<JU 

—                                             •"?'                                       -=r 

\ 

'                                     \ 

<  12                   *• 

/                                      V 

a                                     1)        c                      a 

\                                                            "•'•). 

<                               -                 25'                                               "^ 

i 

<             5'  -       > 

RI 

R2 

FIG.  24. 

cent  rated  at  the  center  of  the  beam.     To  find  the  value  of 
#2  take  moments  about  Ri  thus: 

5X2000+15X1200+18X2400-25^2+30X3000  =  0; 


THEORY  OF  BEAMS  45 

Likewise,  taking  moments  about  R?  gives, 

+5X3000-7X2400-  10Xl200-20X2000+25#i=0; 


PROS.  51.  A  simple  beam  of  25-ft.  span  carries  loads  of  300, 
250,  and  400  Ibs.  distant  5-ft.,  10-ft.,  and  20-ft.  respectively  from 
the  left  end.  Beam  weighs  30  Ibs.  per  lin.  ft.  Find  the  reactions  of 
the  supports. 

PROS.  52.  A  simple  beam  of  20-ft.  span  carries  loads  of  5000, 
3000,  and  4000  Ibs.  located  4-ft.,  10-ft.,  and  12-ft.  from  the  left  end. 
If  the  beam  weighs  35  Ibs.  per  lin.  ft.,  find  the  reaction  of  the  sup- 
ports. 

PROB.  53.  A  beam  28  ft.  long  overhangs  the  right-hand  sup- 
port a  distance  of  7  ft.  The  beam  carries  a  load  of  6000  Ibs.  at 
the  extreme  right  end  and  also  loads  of  2000  and  3000  Ibs.  located 
8  ft.,  and  12  ft.  from  the  left  end.  Find  the  reaction  of  the  supports. 

ART.  19.     BENDING  MOMENT 

It  is  apparent  that  the  tendency  of  the  external  forces,  or 
loads,  acting  on  a  beam  will  be  to  cause  a  flexure  or  bending 
of  the  beam.  This  tendency  increases  with  the  external 
loads  and  the  span  of  the  beam. 

The  bending  moment  at  any  section  of  a  beam  is  defined 
as  the  algebraic  sum  of  the  moments  of  all  the  forces  acting 
to  the  left  of  the  given  section.  The  bending  moment  at 
the  left-hand  support  equals  zero,  as  there  are  no  forces  to 
the  left  of  this  section.  The  bending  moment  at  the  right- 
hand  support  equals  zero  in  order  to  maintain  the  condition 
of  equilibrium.  Between  the  two  reactions  the  bending 
moment  will  vary.  The  point  in  the  beam  at  which  the 
bending  moment  is  the  greatest  or  a  "  maximum  "  is  called 
the  dangerous  section  of  the  beam,  for  at  this  point  the  internal 
stress  is  also  a  maximum.  The  method  of  calculating  this 
bending  moment  is  best  shown  by  a  series  of  typical  prob- 
lems. 


46 


STRENGTH  OF  MATERIALS 


EXAMPLE.  CASE  I.  Weight  of  beam  neglected.  Assume 
a  beam  loaded  as  shown  in  Fig.  25.  By  taking  moments 
about  the  supports  the  reactions  are  found  to  be  Ri  =  2600, 


2406' 


3000 


Fig.  25 


Fig.  26 


Fig.  27 


CASBI 
X 


CASEH 


#2  =  2800.  The  bending  moment  at  the  points  a  =  QXRi  = 
15,600  ft ,-lbs.  The  bending  moment  at  point  b  =  l2Ri  -  6  X 
2400  =  16,800  ft.-lbs.  In  Fig.  26  let  distances  measured 
along  the  vertical  <f>Y  equal  the  bending  moments  to  scale. 


THEORY  OF  BEAMS 


47 


The  line  ocdx  shows  the  variation  in  the  bending  moment  at 
the  various  points  in  the  span  of  the  beam.  To  clearly  show 
this  variation  it  is  usual  to  cross-section  the  area  included 
between  the  curve  and  the  horizontal. 

CASE  II.     Assume  beam  weighs  30  Ibs.  per  lin.  ft.     Each 
reaction  will  be  increased  by  an  amount  equal  to  one-half 


960 


1200 


2400 


3000 


-14- 


-24- 


Fig.  28 


Fig. 


the  total  weight  of  the  beam  or  270  Ibs.,  hence  Ri  =  2870  and 
#2  =  3070. 

The  weight  of  the  section  between  the  point  a  and  the 
reaction  R\  can  be  considered  as  a  single  load  equal  to  30X6 
=  180  Ibs.  concentrated  at  the  center  of  the  section,  assuming 
that  the  beam  is  of  uniform  cross-section.  The  forces  acting 
to  the  left  of  the  point  a  are  the  reaction  Rit  and  the  weight 


48  STRENGTH  OF  MATERIALS 

of  the  6-ft.  section  of  the  beam  concentrated  at  a  distance  of 
3  ft.  from  the  point  a.  The  bending  moment  at  point  a  — 6 
X  #i-190X3  -16,680  ft.-lbs.  The  forces  acting  to  the 
left  of  the  point  b  are  the  reaction  R\,  the  load  of  2400  Ibs. 
and  the  weight  of  the  12-ft.  section  of  the  beam  concen- 
trated at  a  distance  of  6  ft.  from  the  point  6.  Hence  the 
bending  moment  at  point  6  =  12 X #1-6X2400-6X360 
=  17,880  ft.-lbs.  Fig.  27  shows  the  bending-moment  dia- 
gram for  this  case.  Here  it  will  be  noted  the  outline  of  the 
diagram  is  a  curve  rather  than  a  broken  line  as  in  Case  I, 
where  the  weight  of  the  beam  was  neglected. 

EXAMPLE.  A  beam  30  ft.  long  overhangs  the  right-hand 
support  a  distance  of  6  ft.  If  the  beam  is  loaded  as  shown  in 
Fig.  28  and  weighs  40  Ibs.  per  ft.,  find  the  bending  moments 
at  the  points  a,  b,  c,  and  d,  and  draw  the  bending-moment 
diagram. 

SOLUTION.     Taking  moments  about  the  reactions  gives 

#i  =  1820  and  R2  -6940. 

Bending  moment  ata  =  6X#i-6X40X3  =  10200  ft.-lbs. 
Bending  moment  at  b  =  12 X Ri  - 12 X40 X 6  - 960 X 6  =  13200 
ft.-lbs. 

Bending  moment  at  c  =  16X/2i  -16X40X8-960X10- 1200 

X4  =  9600  ft. -Ibs. 

Bending  moment  at  d  =  24X#i  -24X40X12-960X 18- 

1200X12-2400X8-  -18,720. 

The  bending  moments  are  plotted  to  scale  as  shown  in 
Fig.  29. 

PROB.  54.  In  Fig.  23  compute  the  bending  moments  at  the 
points  a,  b,  and  at  the  center  of  the  span.  Draw  a  bending-moment 
diagram. 


THEORY  OF  BEAMS 


49 


PROB.  55.  In  Fig.  24  compute  the  bending  moments  at  the 
points  o,  b,  c,  and  c?,  and  draw  the  bending-moment  diagram. 

PROB.  56.  A  cantilever  beam  8  ft.  long  and  weighing  35  Ibs. 
per  ft.,  carries  a  load  of  4000  Ibs.  6  ft.  from  the  support.  Find  the 
bending  moment  at  the  support  and  at  sections  distant  2,  4,  and  6  ft. 
from  the  support. 


ART.  20.     GENERAL  EQUATION  FOR  BENDING  MOMENT 

All  problems  dealing  with  bending  moments  of  simple 
beams  can  be  solved  by  a  general  rule  which  will  now  be 
formulated.  Assume  a  beam  loaded  as  shown  in  Fig.  30 


oooood 

P3 

p 
a 

1                       ,Y 

QOOCJ 

3 

boood 

boooocbooooooooon 

i               i 

i 

RI 

FIG.  30. 

and  let  y-y  be  any  section  distant  x  feet  from  the  left-hand 
reaction.  Represent  the  distances  from  this  section  to  the 
various  concentrated  loads  P\,  P2,  PS,  by  ai,  a2,  as,  etc.  Let 
w  equal  the  uniformly  distributed  load  in  pounds  per  linear 
foot  plus  the  weight  of  the  beam  in  pounds  per  linear  foot, 
provided  the  weight  is  known. 

The  uniform  load  on  the  x  feet  equals  wx  pounds,  and  this 
may  be  replaced  by  a  single  load  of  wx  pounds  located  at  the 

center  of  the  x  foot  section  or  -  ft.,  from  the  section  y-y. 

The  bending  moment  at  y-y  equals  the  algebraic  sum  of  the 
moments  to  the  left  of  the  section  or  is  equal  to  the  moment 


50 


STRENGTH  OF  MATERIALS 


of  the  left-hand  reaction  minus  the  moment  of  the  concen- 
trated loads,  minus  the  moment  of  the  uniform  load  to  the 
left  of  y-y.  The  moment  of  the  reaction  =  -\-RiX.  The 
moments  of  the  concentrated  loads  =  Piai  —  Pza?.,  —  Pzciz, 
etc.,  briefly  represented  by  —  PSa  (read  summation  of  all  the 
terms  of  the  form  Pa.)  The  moment  of  the  uniformly  dis- 
tributed load  relative  to  the  section 

x        wx2 

y-y=-WX.-= _. 

If  M  equals  the  bending  moment  at  the  section  y^y  it  fol- 
lows that 


WX' 


(8) 


The  application  of  this  general  rule  to  determine  the  maxi- 
mum bending  moment  of  the  four  more  common  types  of 
beams  follow. 

CASE  I.     Simple  beam — concentrated  load  at  the  center, 

p 
weight  neglected,  see  Fig.  31.     Here  Ri  =  -^.    By  inspection 


.                      L                    . 

P 

"^                              2 

i 

1 

1 

i 

\                                                                                                                                 / 

IT' 

i 

FIG.  31. 

it  is  apparent  that  the  bending  moment  will  be  a  maximum 
at  the  center  of  the  span,  therefore  £=«•  and  W  =  o,  since 
the  weight  is  neglected.  Also  2Pa  =  0,  as  there  are  no  con- 


THEORY  OF  BEAMS  51 

centrated  loads  between  the  center  of  the  beam  and  the  left 
reaction.  Substituting  these  values  in  Equation  (8)  there 
results, 


CASE  II.     Simple   beam — uniformly  distributed   load   of 
pounds  per  lin.  ft.  see  Fig.  32.     In  this  case  Ri  =  ^. 

Zi 

QOOOOOOOOOOOOOOOOOO 


FIG.  32. 

The  bending  moment  will  be  a  maximum  at  the  center  of 
the  span,  hence  #  =  —  .     Substituting  these  values  in  Equa- 


tion  (8)  gives 


8  ' 

Let  TF  =  the  total  uniformly  distributed  load  =  wL,  then 

WL 


8  8 


(10) 


(ID 


CASE  III.  Cantilever  beam — concentrated  load  at  the  end, 
weight  neglected,  Fig.  33.  In  the  case  of  cantilever  beams 
the  bending  moment  is  negative  and  becomes  a  maximum  at 
the  point  of  support.  Consider  the  beam  supported  at  the 
right-hand  side.  Then  R\=0,  x  =  L,  and  w  =  o.  Sub- 
stituting these  values  in  the  general  Equation  (8),  there 
results 

M=  -0-PL-0=  -PL  (12) 


52 


STRENGTH  OF  MATERIALS 


CASE  IV.     Cantilever   beam — uniformly   distributed   load 
of  w  pounds  per  foot,  Fig.  34.     Here  Ri  =  0;  x  =  L;  therefore 


WL 


(13) 


FIG.  33. 

NOTE. — It  is  customary  to  represent  concentrated  loads 
by  the  letter  P,  and  the  total  uniformly  distributed  load  by  W. 
The  uniformly  distributed  load  in  pounds  per  linear  foot  is 
represented  by  w. 

OOOOOOOOOOOOQOOOOOOO 


i 

T 

'// 
-1/, 

1 

^                                                       L                                                    ^* 

FIG.  34. 


PROS.  57.  Find  the  maximum  bending  of  a  simple  beam  where 
span  is  18  ft.  The  beam  carries  a  uniform  load  of  200  Ibs.  per  lin.  ft., 
and  a  concentrated  load  of  1800  Ibs.  4  ft.  from  the  left  end. 

PROB  58.  Find  the  maximum  bending  moment  on  a  cantilever 
beam  which  projects  10  ft.  beyond  its  support  and  carries  concen- 
trated loads  of  500  and  1500  Ibs.  located  4  and  8  ft.  from  the  end  of 
the  beam. 


THEORY  OF  BEAMS 


53 


ART.  21.    INTERNAL  RESISTING  MOMENTS 

The  internal  bending  moment  produces  a  tension  in  the 
lower  fibers  of  the  beam  and  a  compression  in  the  upper 
fibers.  As  the  beam  bends  the  upper  fibers  shorten  and  the 
lower  fibers  lengthen.  Somewhere  between  these  two  points 
there  is  a  neutral  plane  which  neither  elongates  nor  shortens. 
It  will  be  shown  later  that  this  neutral  plane  or  axis  passes 
through  the  gravity  axis  of  the  section  of  the  beam. 

It  has  been  proven  experimentally  that  the  stress  on  any 
fiber  of  the  beam  varies  directly  with  its  distance  from  the 
neutral  axis,  provided  the  elastic  limit  of  the  material  is  not 
exceeded . 


1 

P,                / 
<  «-„  >\ 

£       * 

-Oj                 -J 

.*/•"     i 

7          t 

r<                                   V                                     S 

r 

L 

r 

D 

FIG.  35. 

Assume  a  beam  loaded  as  shown  in  Fig.  35.  Let  AB  be 
the  section  where  the  bending  moment  is  a  maximum.  Let 
MN  be  the  neutral  axis.  All  fibers  above  the  axis  MN  will 
be  in  compression  and  all  fibers  below  MN  will  be  in  tension. 
Now  consider  the  right-hand  section  of  the  beam  removed, 
and  a  horizontal  force  of  C  pounds  equal  to  the  total  com- 
pressive  stress  in  top  part  of  beam,  inserted;  also  a  horizontal 
force  of  T  pounds  inserted  to  replace  the  total  tensile  stress 
in  bottom  fibers  of  beam.  It  is  evident  that  the  external 
bending  moment  tends  to  rotate  the  beam  in  a  clockwise 
direction  about  the  point  G.  To  insure  equilibrium  the 


64 


STRENGTH  OF  MATERIALS 


forces  C  and  T  must  produce  a  moment  about  0  equal  and 
opposite  to  the  external  bending  moment.  The  moment 
exerted  by  the  internal  resistance  of  the  beam  is  called  the 
internal  resisting  moment. 

Fig.  36  represents  the  cross-section  of  the  beam  at  AB, 
and  Sc  equals  the  unit  compressive  stress  on  the  outermost 
fiber  of  the  beam  in  compression,  and  St  equals  the  unit 
tensile  stress  on  the  outermost  fiber  of  the  beam  in  tension. 
The  stress  along  the  neutral  line  MN  is  zero.  Take  any 
plane  CD  at  a  distance  of  x  units  from  the  neutral  axis.  Let 
the  thickness  of  this  plane  be  very  small  and  its  cross-sec- 


SECTION  A-B 
FIG.  36. 


tional  area  in  square  inches  be  represented  by  a.  Represent 
the  unit  stress  on  this  area  by  Sx  and  let  the  stress  on  the 
outermost  fiber  be  represented  by  S.  Then 


=       or 

o      c 


The  total  stress  on  the  area  a  will  equal  the  unit  stress 
( S  X  - )  tunes  the  area  a,  or  equal  ( —  X  ax  j .  This  total  stress 

on  the  area  a  exerts  a  resisting  moment  about  the  point  0 
which  tends  to  counteract  the  external  bending  moment. 
The  resisting  moment  of  the  resistance  of  the  area  a  equals 


THEORY  OF  BEAMS  55 

or 

the  total  stress  —ax  times  the  moment  arm  x,  or  equals 
c 

S  S 

—  ax-x  =  —  Xax2. 

c  c 

Now,  consider  the  total  cross-section  of  the  beam  at  AB  to 
be  made  up  of  an  infinite  number  of  small  areas  of  the  form 
a.  Let  N  equal  the  number  of  these  areas  each  equal  to  a 
in  value  and  let  A  equal  the  total  cross-section  of  the  beam. 
Then  it  is  evident  that  A  =  aN.  The  total  resisting  moment 
of  the  beam  will  equal  the  sum  of  the  resisting  moments  of 
each  elementary  area,  or  the  resisting  moment  equals 

—  axi2-\  —  ax22-\  —  axz2   .    .    -\  —  axn2 
c  c  c  c 


.    .    +x*\.     .     (14) 
Let 

(X!2  +  X22  +    .    .    .    +  Xn2)=NX2, 

where  X2  =  the  average  of  all  the  terms  in  this  sum.  Sub- 
stituting this  value  in  Equation  (14)  there  results,  the  resist- 
ing moment  equals 


then  since  there  are  N  terms  it  follows  that 

.  .  .  +xn2)=NX2; 


but  aN  =  A  (the  total  cross-section  of  the  beam)  ;  hence  the 

or 

resisting  moment  =  —  (AX2).     It  is  evident  that  the  term 

AX2  is  entirely  dependent  on  the  form  of  cross-section  of  the 
beam.  This  factor  might  be  called  "  the  factor  of  strength," 
but  is  commonly  referred  to  as  the  moment  of  inertia  of 


56  STRENGTH  OF  MATERIALS 

the  section  and  is  generally  represented  by  7;  or  the  resisting 
moment  equals 

?  ........    (15) 

OT 

For  equilibrium  the  internal  resisting  moment  --  must 

c 

equal  the  external  bending  moment  M,  or  there  results  the 
fundamental  equation  for  the  design  of  beams 

ar 

M=^  .......     (16) 

This  equation  can  be  written, 


The  term  f  —  )  is  called  the  "  Section  Modulus"  of  the  beam. 


It  is  to  be  noted  that  the  section  modulus  is  entirely  inde- 
pendent of  the  material  of  which  the  beam  is  made  and 
depends  only  upon  the  distribution  of  the  metal  relative 
to  the  neutral  axis  of  the  section. 

PROB.  59.  Consider  a  rectangular  beam  of  breadth  6  ins.  and 
depth  d  ins.  Divide  the  cross-section  into  20  rectangles  of  breadth 

b  ins.  and  depth  —  in.     Derive  an  expression   for   the    resisting 

moment  n  terms  of  b,  d,  and  the  maximum  fiber  stress  S. 

PROB.  60.  Find  the  total  compressive  stress  and  the  total 
tensile  stress  on  a  rectangular  beam  whose  depth  is  12  ins.  and 
breadth  8  ins.  Stress  on  outer  fiber  equals  1000  Ibs.  At  what 
point  in  the  section  is  the  stress  equal  to  zero? 

ART.  22.     CENTER  OF  GRAVITY 

A  body  may  be  considered  as  made  up  of  a  series  of  small 
particles.  The  weights  of  all  these  particles  form  a  system  of 
vertical  parallel  forces,  and  the  resultant  of  this  system  must 


THEORY  OF  BEAMS  57 

evidently  equal  the  sum  of  these  forces  or  the  total  weight  of 
the  body.  The  point  at  which  this  resultant  force  or  weight 
acts  is  called  the  center  of  gravity  of  the  body.  As  commonly 
defined  the  center  of  gravity  of  a  body  is  the  point  through 
which  the  line  of  action  of  the  weight  of  the  body  always 
passes. 

In  the  following  discussion  the  term  center  of  gravity  will 
be  abbreviated  to  the  symbol  e.g.  If  a  bar  of  uniform  cross- 
section  be  suspended  by  a  rope  attached  at  one  end  of  the  bar, 
it  is  evident  that  the  bar  will  hang  in  a  vertical,  and  not  in  a 
horizontal,  position;  or,  if  the  rope  is  attached  to  the  center, 
the  bar  will  probably  assume  a  horizontal  position.  Hence  a 
vertical  line  drawn  through  the  point  of  suspension  of  a  body 
must  always  pass  through  the  e.g.  of  the  body. 

The  e.g.  of  bodies  which  are  symmetrical  with  respect  to 
a  given  point,  and  are  of  uniform  density,  will  be  at  the  given 
point.  Thus  the  e.g.  of  a  sphere  is  at  its  geometrical  center 
and  the  e.g.  of  a  circle  is  at  its  center.  The  student  can 
readily  think  of  many  more  examples.  Therefore,  in  many 
cases  the  determination  of  the  e.g.  is  simply  a  matter  of 
inspection.  In  the  case  of  unsymmetrical  figures  the  e.g. 
may  be  found  experimentally  by  making  a  template  of  card- 
board to  represent  the  body  to  a  given  scale.  Next  suspend 
the  template,  first  from  one  point  and  then  from  another, 
and  in  each  case  draw  a  vertical  line  through  the  point  of 
suspension.  The  intersection  of  these  two  vertical  lines  will 
locate  the  e.g.  of  the  template,  from  which  the  e.g.  of  the 
body  can  readily  be  found.  Take,  for  example,  the  counter- 
weight attached  to  the  crank  of  certain  forms  of  steam  en- 
gines. Let  the  weight  be  of  uniform  thickness  and  its  cross- 
section  be  of  the  form  shown  in  Fig.  37,  which  is  a  template 
of  the  given  weight.  If  the  template  be  supported  from  the 
point  0  and  the  vertical  line  CD  be  drawn  through  the  point 
of  suspension,  then  the  e.g.  of  the  body  will  lie  somewhere  on 
this  line.  Likewise,  if  the  template  be  suspended  from  the 


58 


STRENGTH  OF  MATERIALS 


point  B  the  e.g.  will  lie  on  the  line  AB.     Therefore,  the  e.g. 
of  the  body  will  be  found  at  the  point  E  of  intersection  of  the 

two  lines  CD  and  AB.  This 
method  is  frequently  used  in 
the  drafting  room,  as  no  com- 
putations are  necessary. 

If  a  body  can  be  divided  into 
regular  geometrical  figures,  such 
as  triangles,  squares,  and  rect- 
angles, the  e.g.  is  best  obtained 
by  applying  the  law  that  the  mo- 
ment of  the  resultant  (which  is 
the  total  weight  of  the  body)  is 
equal  to  the  sum  of  the  moments 
of  the  other  forces  (the  weights  of 
each  of  the  separate  figures). 

In  problems  where  the  body  is  symmetrical  with  neither 
the  vertical  nor  the  horizontal  axes  the  following  general 
method  must  be  applied  to  find  the  e.g.  Assume  a  given  body 
to  be  divided  into  a  large  number  of  very  small  parts,  and  the 


FIG.  37. 


FIG.  38. 

weights  of  each  part  to  be  represented  by  the  points  wi,  W2,  WB, 
etc.,  as  shown  in  Fig.  38.  Locate  each  of  these  weights  from 
the  axes  OX  and  OF,  thus  w\  is  x\  units  from  the  OY  axis, 
and  yi  units  from  the  OX  axis.  The  resultant  of  all  these 


THEORY  OF  BEAMS  59 

small  weights  will  be  the  total  weight  W  of  the  given  body. 
Thus  W  =  wi,  W2,  wz  .  .  .,  etc.  This  weight  or  resultant 
will  act  at  the  e.g.  of  the  body.  Let  this  point  be  X  units 
from  the  OY  axis  and  Y  units  from  the  OX  axis.  Then  the 
distance  Y  may  be  found  by  taking  moments  about  the  axis 
OX. 

Thus, 


or 

v_ 
•*•  — 


W1+W 


In  like  manner  the  value  of  X  may  be  found  by  taking 
moments  about  the  OY  axis.     Thus, 


V_  /1nx 

.A  —  i       ;     ~~j  -  .       .    .    .    (  iy) 


Fig.  39  shows  the  cross-section  of  a  riveter  frame.  Locate 
the  e.g.  of  this  section.  In  dealing  with  problems  of  this 
kind  the  moments  of  the  areas  are  taken  to  determine  the 
location  of  the  e.g.,  hence  Equations  (19)  and  (18)  can  be 
written 


This  figure  is  symmetrical  about  the  vertical  axis  and, 
hence,  its  e.g.  lies  somewhere  on  the  line  AB  say  at  the  point 
H,  x  inches  from  the  line  CD.  Divide  the  section  into  the 
three  rectangles  I,  II,  and  III.  By  inspection  it  is  clear 
that  the  e.g.  of  each  rectangle  is  at  the  point  of  intersection 
of  the  diagonals,  and  also  that  the  weight  of  each  part  is 
proportional  to  the  area  of  the  cross-section.  The  moment 
of  the  weight  of  each  section  about  any  given  point  will  then 
be  proportional  to  the  moment  of  the  area, 


60 


STRENGTH  OF  MATERIALS 


The  e.g.  of  part  III  is  located  9J  ins.  from  the  line  CD,  of 
part  II,  5}  ins.  from  CD,  and  part  I,  f  in.  from  CD.  The 
area  of  part  I  equals  5jXli  =  8j  sq.  in.,  the  area  of  part  II 
equals  7JXi  =  3f  sq.  in.,  and  the  area  of  part  III  equals 


FIG.  39. 


2X1  =  2  sq.  in.  Now,  take  moments  of  the  areas  I,  II  and 
III,  about  the  line  CD.  The  sum  of  these  moments  must 
equal  the  moment  of  the  resultant  (total  area) .  Thus, 


or 


x  =  3.2  in. 


THEORY  OF  BEAMS 


61 


While  the  above  method  is  accurate,  errors  are  hard  to 
locate  if  the  equations  contain  many  terms.  In  order  to 
readily  check  results  it  is  desirable  to  tabulate  the  data  and 
results  as  shown  in  Table  IX. 

The  e.g.  of  the  cross-section  of  any  compound  shape 
can  quickly  be  located  by  this  scheme.  Divide  the  section 
into  rectangles  and  triangles;  locate  the  e.g.  of  each  section 
with  respect  to  the  base  of  the  figure.  Mark  each  section, 
give  its  dimensions,  area,  moment  arm,  and  moment.  The 
sum  of  the  moment  column  (5)  in  Table  IX  divided  by  the 
total  area  as  given  in  column  (3)  will  give  the  distance  X 
from  the  base  of  the  figure  to  its  e.g. 

TABLE  IX 
COMPUTATIONS  FOR  CENTER  OF  GRAVITY 


1 

2 

3 

4 

5 

Mark. 

Dimensions, 
Inches. 

Area, 
Sq.  In. 

Moment 
Arm,  In. 

Moment. 

I 

5|XH 

8| 

1 

4 

6.18 

II 

7*X  \ 

3| 

51 

19.70 

III 

2   X  1 

2 

91 

19.00 

14 

44.88 

X  = 

3.2 

The  elementary  areas  or  weights  are  in  equilibrium  with 
respect  to  the  gravity  axis.  In  Equation  (20)  X  =  0,  with 
reference  to  the  gravity  axis,  and  it  follows  that  0,1X2+ 
(12x2+0,3X3+  ...  =0  or  2ax  =  Q.  That  is,  the  summation 
of  each  elementary  area  times  its  distance  from  the  gravity 
axis  equals  0. 

In  Fig.  36  the  total  stress  on  any  element  of  area  equals 

S 

a-x--.     The  line  MN  represents  the  neutral  axis.     From 
c 

the  conditions  of  equilibrium  it  is  evident  that  the  sum  of 


62  STRENGTH  OF  MATERIALS 

the  tensile  stresses  below  the  neutral  axis  must  equal  the 
sum  of  the  compressive  stresses  above  the  neutral  axis,  and 
that  the  algebraic  sum  of  all  the  stresses  relative  to  the  neu- 
tral axis  equals  0.  The  summation  of  the  stresses  on  all  the 

S 

elements  of  area  equals  -Saz  =  0,  hence  Zaz  =  0,  therefore  the 
c 

neutral  axis  of  any  beam  is  coincident  with  the  gravity  axis 
of  the  section. 

PROB.  61.  Locate  the  gravity  axis  of  the  vertical  section  of  the 
150-ton  punch  frame  shown  in  Fig.  46. 

PROB.  62.  Locate  the  gravity  axis  parallel  to  the  longer  leg 
of  a  5x4-in.  angle;  thickness  of  each  leg  equals  \  in. 

PROB.  63.  Locate  the  gravity  axis  of  a  12-in.  channel  section; 
average  thickness  of  flange  equals  0.5  in.;  thickness  of  the  web 
equals  .28.  Depth  of  flange  2.94  in. 

PROB.  64.  Locate  the  gravity  axis  of  a  T  bar,  whose  depth  is 
4  ins.,  width  of  flange  4  ins.,  average  thickness  of  flange  f  in.; 
average  thickness  of  stem  equals  f  in. 


ART.  23.    MOMENT  OF  INERTIA 

In  the  formula  for  the  internal  resisting  moment  of  a 
beam  the  factor  I  —  2ax2,  is  called  the  moment  of  inertia  of 
the  section,  but  as  stated  before,  this  term  should  be  con- 
sidered only  as  a  "  factor  of  strength."  Its  numerical  value 
depends  on  the  distribution  of  the  metal  in  the  beam  relative 
to  the  neutral  axis. 

The  moment  of  inertia  is  the  quantity  obtained  by  mul- 
tiplying each  elementary  area  of  a  given  section  by  the  square 
of  its  distance  from  the  axis  about  which  the  moment  is 
desired.  Moment  of  inertia  of  various  sections  can  be 
added  and  subtracted  provided  they  are  figured  about  a 
common  axis.  To  determine  accurately  an  expression  for 
the  moment  of  inertia  of  the  common  sections  such  as  rect- 


THEORY  OF  BEAMS 


63 


angles,  squares,  triangles,  etc.,  involves  the  use  of  the  cal- 
culus, so  in  this  text  no  such  values  will  be  derived,  but 
simply  stated;  thus  the  moment  of  inertia  of. a  rectangle 

bd3 
about  its  gravity  axis  equals  —=-,  where  b  equals  the  breadth 

in  inches,  and  d  equals  the  depth  in  inches.     The  moment  of 

bh3 
inertia  of  a  triangle  about  its  gravity  axis  equals  ^-,  where 

oO 

6  equals  the  base  and  h  equals  the  altitude.     For  a  circle 

T       Trd4       _,  T        (P 

g==~Q4'  square  Ig  =  ^. 

In  many  ceases  it  is  desirable  to  transfer  the  moment  of 
inertia  from  the  gravity  axis  to  some  other  axis  which  is 


FIG.  40. 

parallel  to  the  given  axis.  The  following  formula  shows  the 
relation  between  the  moment  of  inertia  about  the  gravity 
axis  g-g  (see  Fig.  40),  and  any  axis  (x-x)  parallel  to  the 
gravity  axis. 

Ix  =  Ig+Ad2, (22) 

where 

Ix  =  Moment  of  inertia  about  given  axis; 

I0  =  Moment  of  inertia  about  gravity  axis; 
A  =  the  area  of  the  section  in  square  inches ; 
d  =  distance  in  inches  between  the  parallel  axes. 


64 


STRENGTH  OF  MATERIALS 


One  of  the  more  common  applications  of  this  reduction 
formula  is  in  the  case  of  the  rectangle.     Thus  in  Fig.  40  the 

value  of  Ig  =  —  0  ;    A  =  bh  ;    d  =  ~  ;    Ix  =  moment  of  inertia 

i  _  z 

about  the  base,  hence 

h2    bh* 


For  a  hollow  rectangle  Ig  =  -^  —  1—~-     (See  Fig.  41.) 


For  the  I  section  Ig  =  -7r- 

o 


.     (Fig.  42.) 


FIG.  41. 


FIG.  42. 


In  any  section  which  is  not  symmetrical  with  respect  to 
the  gravity  axis,  the  moment  of  inertia  can  readily  be  found 
by  dividing  the  section  into  rectangles  located  above  and 
below  the  gravity  axis.  In  each  case  the  base  of  these 
elementary  rectangles  must  be  coincident  with  the  gravity 
axis  of  the  figure.  For  the  T  section  the  values  of  c  and  c\ 
must  be  determined  as  explained  in  Art.  22.  In  Fig.  43 
the  moment  of  inertia  of  the  T  section  is  found  by  taking  the 
moment  of  inertia  of  the  part  below  the  gravity  axis,  which 


equals 


and  adding  the  moment  of  inertia  of  the  part 


THEORY  OF  BEAMS  65 

503       (b  —  t i)   (c  —  t}3 

above  the  gravity  axis,  which  equals   -^ — —~ —  — . 

The  value  -^-  is  used  because  the  base  of  the  rectangle  is 

coincident  with  the  gravity  axis  of  the  entire  figure.     Hence 
the  moment  of  inertia  of  a  T  section  equals 


(b-t  )(c-Q8 


FIG.  43. 


FIG.  44. 


For  a  channel  section  (see  Fig.  44)  whose  breadth  is  b,  depth 
h,  thickness  of  flange  t,  thickness  of  web  ti,  the  value  of 


For  the  angle  (Fig.  45)  whose  depth  is  h,  thickness  leg  t,  the 
value  of 

(h-t)(ci-t)* 


These  values  are  correct  in  the  case  of  cast-iron  section. 
Standard  steel  shapes  do  not  have  parallel  edges,  hence  the 
above  formula?  do  not  hold.  For  exact  values  the  student 
is  referred  to  handbooks  on  structural  steel. 

For  compound  sections  the  value  of  Ig  is  determined  by 
dividing  the  section  into  a  series  of  rectangles  and  finding 
the  value  of  the  moment  of  inertia  of  each  rectangle  with 


66 


STRENGTH  OF  MATERIALS 


respect  to  the  gravity  axis  of  the  figure.  It  is  advisable  to 
use  a  standard  form  for  tabulating  results  as  shown  in 
Table  X;  assuming  that  the  gravity  axis  has  been  located 
as  explained  in  Art.  22. 

To  use  this  method  the  figure  whose  moment  of  inertia 
is  to  be  found,  is  divided  into  a  series  of  rectangles,  each  of 
which  has  its  base  coincident  with  the  gravity  axis  of  the 
figure.  These  rectangles  are  marked  I,  II,  III,  etc.  These 
marks  are  placed  in  column  1,  Table  X.  Column  2  gives 
the  breadth  of  the  rectangle.  Rectangle  I  is  considered  as 
the  difference  of  two  rectangles,  each  having  a  breadth  of 


Table  X. 


14  ins.  and  depths  of  13  ins.  and  11  ins.  respectively.  Column 
3  gives  the  greater  depth,  designated  h\\  and  column  4  gives 
the  lesser  depth  designated  /?2-  Column  5  gives  the  values  of 
hi3.  These  values  are  quickly  determined  by  the  use  of  a 
table  of  cubes  of  numbers.  Column  7  gives  the  values  of 
(hf—ho3).  Column  8  gives  the  values  of  bX(hi3  — h£). 
The  values  in  column  8  will  be  three  times  the  moment 
of  inertia  of  the  given  rectangle,  as  the  moment  of  inertia 
of  a  rectangle  about  its  base  is  6/i3;  hence  the  final  total  of 
column  8  is  divided  by  3  to  give  the  moment  of  inertia  of  the 


THEORY  OF  BEAMS 


67 


entire  figure.     This  method  eliminates  the  necessity  of  using 
the  transformation  formula. 

TABLE  X 


1 

Mark 

2 
b 

hi 

4 

/>2 

5 
hi* 

6 
fti» 

7 
(h,*-h& 

8 
6X(/u3—  fo3) 

I 

14 

13 

11 

2,197 

1,331 

866 

12,124 

II 

2 

11 

0 

1,331 

0 

1,331 

2,662 

III 

2 

13 

0 

2,197 

0 

2,197 

4,394 

IV 

10 

15 

13 

3,375 

2,197 

1,178 

11,780 

3      30,960 

Ig- 

10,320 

PEOB.  65.     Find  the  moment  of  inertia  about  the  gravity  axis 
for  the  section  shown  in  Fig  46. 


FIG.  46. 


FIG.  47. 


PROB.  66.  A  plate  girder  is  constructed  as  shown  in  Fig.  47. 
Find  the  moment  of  inertia  of  the  section  with  respect  to  the  gravity 
axis. 

PROB.  67.  (a)  Find  the  moment  of  inertia  about  the  gravity 
axis  of  a  circle  3  ins.  in  diameter,  (b)  Find  the  moment  of  inertia 
of  the  circle  about  an  axis  parallel  to  and  located  8  ins.  from  the 
gravity  axis. 


68  STRENGTH  OF  MATERIALS 

ART.  24.     REVIEW  PROBLEMS 

PROB.  68.  Locate  the  gravity  axis  of  a  standard  12-in. 
channel  weighing  20.5  Ibs.  per  foot.  For  dimensions  of  section 
see  Cambria  handbook. 

PROB.  69.  A  beam  30  ft.  in  length  overhangs  each  support. 
The  left-hand  support  is  located  4  ft.  to  the  right  of  the  extreme 
left  end  of  the  beam.  Distance  between  supports  is  20  ft.  Beam 
carries  concentrated  loads  of  2000,  3000,  4000,  and  2000  Ibs.  located 
0,  5,  10,  15,  and  30  ft.  from  the 'left  end  respectively.  Draw  the 
bending-moment  diagram. 

PROB.  70.  Fig.  48  shows  a  cast-iron  bracket.  Find  the  moment 
of  inertia  of  the  section  AB  with  respect  to  the  gravity  axis. 


SECTHON  A-B 


FIG.  48. 


PROB.  71.  Find  the  moment  of  inertia  relative  to  the  gravity 
axis  of  the  box  girder  shown  in  Fig.  49,  if  the  plates  are  14  ins.  in 
width. 


_la"l  Beam 
80  Ib.  per  ft 


FIG.  49. 


THEORY  OF  BEAMS 


69 


PROB.  72.  Fig.  50  shows  the  cross-section  of  a  crane  hook. 
Locate  the  gravity  axis  of  the  section  and  find  the  moment  of 
inertia  of  the  section  relative  to  the  gravity  axis. 


FIG.  50. 


FIG.  51. 


PROB.  73.  Fig.  51  shows  the  cross-section  of  a  riveter  frame. 
Find  the  moment  of  inertia  of  the  section  relative  to  the  gravity 
axis,  g-g. 

PROB.  74.  A  simple  beam  is  loaded  as  shown  in  Fig.  52.  Draw 
the  bending-moment  diagram. 


iuu  IDS.  per  it. 

^                     fi  '               ^ 

,     ylOOlbs.  perf 

'1 

> 

FIG.  52. 

PROB.  75.    Find  the  moment  of  inertia  of  an  8xl2-in.  wooden 
beam,  about  its  horizontal  and  vertical  gravity  axes. 


CHAPTER  V 
DESIGN  OF  BEAMS 

ART.  25.     SAFE  LOADS — WOODEN  BEAMS 

IN  designing  beams,  especially  when  used  for  the  sup- 
port of  floors,  two  kinds  of  loads  are  recognized.  Live  loads 
consist  of  the  weight  of  carriages,  cranes,  or  other  handling 
devices  and  their  supported  loads,  machinery,  merchandise, 
persons,  or  other  moving  objects.  Dead  loads  consist  of 
the  actual  weight  of  the  structure  itself  with  the  walls, 
floors,  partitions,  roofs,  and  all  other  permanent  fixtures. 

Wooden  beams  are  used  for  the  support  of  floors;  they 
are  generally  rectangular  in  shape  and  supported  at  the  ends 
by  either  steel  or  wood  girders.  The  spacing  of  the  floor 
beams  depends  upon  the  span  and  the  load  carried  in 
pounds  per  square  foot  by  the  floor.  This  load  is  fixed  by 
the  character  of  the  building.  The  safe  unit  working  stress 
for  wooden  beams  is  relatively  low,  owing  to  the  necessity 
of  using  a  high  factor  of  safety  on  account  of  the  non-uniform 
structure  of  the  wood. 

Me 
The  formula  S=  — jr-  is  the  basis  for  determining  the  safe 

load  on  any  beam.  For  wooden  beams  of  rectangular  cross- 
section  this  formula  may  be  stated  in  terms  of  the  external 
load,  the  breadth  and  depth  of  the  beam,  and  the  span. 
Consider  the  case  of  a  simple  wooden  beam  carrying  a  con- 
centrated load  of  P  pounds  at  the  center  of  a  span  of  L  inches. 
Let  b  equal  the  breadth  and  d  equal  the  depth  of  the  cross- 

70 


DESIGN  OF  BEAMS  71 

section  of  the  beam.       From  Equation  (9)  the  maximum 

bending  moment  is  -.-;   the  moment  of  inertia  I  =  ^r,  and 
4  \2t 

c  =  o  (grayity  axis  being  coincident  with  neutral  axis  at  the 

A 

center  of  the  section).     Substituting  these  values  in  equa- 
tion (16)  gives 

bd* 
PL          12 


or 


and 


P=2-g^, (23) 


S---—  (24) 

~2  bd2' 


In  Equations  (23)  and  (24)  the  values  of  b,  d,  and  L  are 
to  be  expressed  in  inches.  Equation  (23)  shows  that  the 
load  varies  as  the  breadth  of  the  beam  and  as  the  square  of 
the  depth,  which  is  the  reason  for  usually  having  the  depth 
of  a  beam  greater  than  its  breadth.  For  example,  assume 
the  case  of  a  yellow  pine  floor  beam  3X8  ins.  in  cross-section 
and  12-ft.  span.  Let  S=  1000  Ibs.  per  sq.  in.;  then  6  =  3  in., 
d  =  8  in.,  S=  1000,  and  L=  12  ft.  =  144  ins.,  hence 


Upon  reference  to  structural  hand  books  it  will  be  noted 
that  there  is  a  fixed  maximum  and  minimum  span  for  wooden 
beams  of  any  given  depth. 

If  the  load  is  uniformly  distributed  the  maximum  bending 


72  STRENGTH  OF  MATERIALS 

WL 
moment  equals  —  ^-.     Substituting  this  value  in  Equation 

(16)   gives 

Sbd* 


___  ( 

8   "      d  3L  ' 

2 
and 


Im  the  previous  example,  if  the  load  is  uniformly  distributed 
instead  of  concentrated  at  the  center  of  the  span,  the  load  is 

4S6d2    4X1000X3X64 

SIT         ^XT44~     =178°lbs- 

The  floors  of  cotton  mills  can  generally  be  light,  for  the 
total  weight  of  machinery,  men,  and  materials  will  seldom 
exceed  30  Ibs.  per  sq.  ft.  Rooms  for  pattern  storage  rarely 
carry  more  than  150  Ibs.  per  sq.  ft.  Buildings  for  light 
machinery  frequently  have  provision  for  loads  of  250  to  300 
Ibs.  per  sq.  ft. 

*  Table  XI  gives  the  safe  working  stresses  for  various  kinds 
of  timber,  based  upon  data  recommended  by  the  American 
Railway  Association. 

PROB.  76.  A  wooden  beam  8x12  ins.  in  cross-section  and 
having  a  span  of  16  ft.  carries  a  uniformly  distributed  load  of  8000 
Ibs.  Find  the  unit  fiber  stress  and  the  factor  of  safety. 

PROB.  77.  A  floor  is  supported  by  wooden  beams  3x8  ins.  in 
cross-section,  having  a  span.of  12  ft.  If  the  beams  are  spaced  18  ins. 
center  to  center,  find  the  safe  load  that  can  be  placed  on  the  floor  in 
pounds  per  square  foot. 

PROB.  78.  A  wooden  beam  having  a  span  of  16  ft.,  carries  a 
concentrated  load  of  2160  Ibs.  at  a  point  10  ft.  from  the  left  end  of 
the  beam.  Assuming  that  the  depth  of  the  beam  equals  1|  the 
breadth,  find  the  dimensions  of  beam,  using  a  safe  working  stress  of 
1200  Ibs.  per  sq.  in. 


DESIGN  OF  BEAMS 


73 


10       '.   O       '.   «0   O       '.   O   O       .' 


oioeoo     •  o  o     •  o  o  o 


s 


C^>    CO   <M   <M 


!8is 

•^     r-H     TT1     i-H 


o  o     •  o  o  o 

O    CO       •    iO    lO    iO 


888 


O 


74  STRENGTH  OF  MATERIALS 

ART.  26.    SHEAR  DIAGRAMS 

The  vertical  shear  at  any  point  in  a  beam  is  equal  to  the 
algebraic  sum  of  all  the  forces  acting  to  the  left  of  the  section. 
The  vertical  shear  will  be  the  maximum  at  one  of  the  reac- 
tions of  the  beam.  Therefore  the  internal  stress  at  the 
reaction  is  one  of  pure  shear.  The  cross-section  of  the  beam 
at  this  point  must  be  such  thai;  the  unit  shearing  stress  shall 
be  great  enough  to  insure  the  proper  factor  of  safety.  Let 
V  equal  the  maximum  vertical  shear  in  pounds;  A  equal 
the  cross-section  of  the  beam  in  square  inches  and  S  equal 
the  safe  working  unit  shearing  stress,  then: 

V  =  AS,        ......     (27) 

or 

S  =  J  ........     (28) 

It  is  customary  to  figure  the  vertical  shear  at  the  reactions, 
under  each  concentrated  load  and  at  the  center  of  the  beam. 
These  values  are  then  plotted  to  scale  forming  a  "  shear 
diagram."  The  vertical  shear  at  any  point  in  a  bea*m  dis- 
tant x  feet  from  the  left  reaction  is  given  by  the  equation. 


(29) 


where  R\  equals  the  left-end  reaction  in  pounds;  SP  equals 
the  algebraic  sum  of  the  concentrated  loads  to  the  left  of  the 
section,  and  wx  equals  the  uniformly  distributed  load  includ- 
ing the  weight  of  the  beam. 

EXAMPLE.  Given  a  beam  loaded  as  shown  in  Fig.  53. 
Plot  the  shear  diagram. 

SOLUTION.     Taking  moments  about  R\  gives 

5X1000+10X2000+15X2400-25^2  =  0, 
or 

#2  =  2440. 

In  like  manner  taking  moments  about  #2  gives  R\  =  2960. 


DESIGN  OF  BEAMS 


75 


The  vertical  shear  at  a  =  R\  =  +2960.  Consider  a  point 
a  very  short  distance  to  the  left  of  the  point  b.  The  shear 
at  this  equals  R,  since  there  are  no  vertical  loads  between 
points  a  and  b  (the  weight  of  the  .beam  is  neglected  in  this 
case).  Now  consider  a  section  a  short  distance  to  the  right 
of  the  point  6;  the  shear  at  this  point  will  be  (Ri  — 1000)  or 
1960  Ibs.  It  is  customary  to  assume  that  the  value  of  the 


2960 


-2440 


FIG.  53. 


vertical  shear  changes  at  each  concentrated  load.     Hence 
the  shear  at  point 

c  =  Ri  - 1000  -  2000  =  2960  -  3000  =  -  40  Ibs. 

It  will  be  noted  that  at  the  point  c  the  vertical  shear 
changes  from  a  positive  to  a  negative  value;  the  shear  at 
point  d  =  #1-1000 -2000 -2400  =-2400;  the  shear  at 
R2=  -2440  Ibs.  The  curve  OABCDEFGH,  Fig.  53,  shows 
at  a  glance  the  variation  in  the  vertical  shear. 


76 


STRENGTH  OF  MATERIALS 


In  cases  where  both  uniform  and  concentrated  loads  are 
involved  the  curve  will  be  of  a  different  form.  In  Fig.  54 
the  beam  carries  a  uniformly  distributed  load  of  200  Ibs. 
per  lin.  ft.,  in  addition  to  the  concentrated  loads  indicated. 
By  taking  moments  about  Ri  and  Tfe  the  reactions  are 
found  to  be  #i=4600  and  fl2=4700  Ibs.  To  find  the  ver- 


800  1600 

-•— J 

<—  4— H 

-)QQQQVQOOOOOOO#) 


2100 


^oooooQoorj 


-21- 


FIG.  54. 

tical  shear  at  the  various  points  substitute  in  Equation  (28) 
the  various  values,  thus  at  point 
a,  7  =  #i-0-0  =  4600  (herez  =  0); 

6,  7  =  fli -800 -200Xz  =  4600 -1600  =  3000  (herez  =  4); 
C)    y  =  JR1-800-1600-200XlO  =  4600-4400  =  200  (here  x 

=  10); 

d,  7  =  7^1-800-1600-2100-200X16  =  4600-7700  = 
-3100  (here  x=  16). 

The  curve  OABCDEFGH,  Fig.  54,  shows  the  variation  of  the 
vertical  shear  in  this  case. 


DESIGN  OF  BEAMS  77 

The  shear  diagram  is  helpful  in  locating  the  dangerous 
section  of  a  beam,  for  it  can  be  proven  that  the  bending 
moment  is  always  a  maximum  at  the  point  where  the  vertical 
shear  passes  through  its  zero  value.  In  Fig.  53  it  will  be 
noted  that  the  shear  passes  from  a  positive  to  a  negative 
value  at  the  point  c,  and,  therefore,  the  bending  moment 
will  be  a  maximum  at  this  point.  The  curve  OMNPS, 
Fig.  53,  shows  the  bending-moment  diagram. 

As  a  general  rule  beams  will  fail  due  to  the  internal  stress 
caused  by  bending,  hence  if  a  beam  is  strong  enough  to 
withstand  the  bending  action  the  vertical  shear  will  take 
care  of  itself.  However,  it  is  well  to  investigate  the  unit 
vertical  shear  at  the  reactions.  For  example,  in  Fig.  53 
the  beam  is  6X9  =  54  sq.  ins.  in  cross-section.  The  vertical 

V    2960 

shear  at  72 1  =  2960,  therefore  the  unit  shear  = -j  = -^- =  55 

A.        O4 

3000 

Ibs.,  and  the  factor  of  safety  in  shear  =  -^=-  =55. 

oo 

In  wooden  beams  of  relatively  large  depth  compared 
with  breadth  the  question  of  horizontal  shear  must  be  in- 
vestigated. A  proper  analysis  of  horizontal  shear  is  beyond 
the  scope  of  this  text,  but  it  may  be  stated  that  for  beams  of 
rectangular  cross-section  the  horizontal  shear  equals  f 
the  vertical  shear. 

PROB.  79.  In  Fig.  54  find  the  maximum  vertical  shear  in  pounds 
per  square  inch,  assuming  the  cross-section  to  be  8X10. 

PROB.  80.  Draw  the  vertical  shear  diagram  for  the  beam 
shown  in  Fig.  28,  and  compare  with  the  bending-moment  diagram 
shown  in  Fig.  29.  Is  the  bending  moment  a  maximum  at  point 
where  the  vertical  shear  passes  through  zero? 

PROB.  81.  Assume  data  given  in  Fig.  54.  Locate  the  danger- 
ous section  of  the  beam  by  use  of  the  law  that  the  bending  moment  is 
a  maximum  at  the  point  where  the  vertical  shear  passes  through 
zero. 


78  STRENGTH  OF  MATERIALS 


ART.  27.     SAFE  LOADS — STEEL  BEAMS 

Wooden  beams  are  being  used  less  and  less  in  building 
construction,  owing  to  the  fact  that  they  do  not  produce  a 
fireproof  structure. 

Steel  beams  are  made  of  open-hearth  steel  having  a 
tensile  strength  of  60,000  to  70,000  Ibs.  per  sq.  in.  and  an 
elongation  of  25  per  cent.  The  steel  is  first  cast  into  large 
ingots,  which  are  then  placed  into  "  soaking  "  pits,  where 
they  are  heated  to  a  high  temperature.  The  ingots  are  taken 
from  these  pits  by  overhead  cranes  and  passed  to  the  rolls, 
where  they  are  formed  into  any  desired  shape.  The  most 
common  shape  is  the  I  section,  which  gives  a  large  moment 
of  inertia  per  pound  weight  of  beam.  The  various  steel 
companies  roll  these  beams  into  standard  shapes  and  weights. 
Table  XII  gives  the  properties  of  the  light  and  heavy  I 
beams  for  various  depths.  On  the  larger  sizes  there  are  two 
and  three  intermediate  weights.  For  properties  of  these 
sections  the  student  is  referred  to  the  handbooks  prepared 
by  the  various  steel  companies. 

Fig.  55  shows  the  method  of  increasing  the  weight  of 
various  rolled  shapes.  For  a  given  depth  the  weight  of  the 
I  beam  is  increased  by  increasing  the  thickness  of  the  web, 
which  necessarily  increases  the  width  of  the  flange.  Column 
6,  Table  XII,  gives  the  moment  of  inertia  of  the  section 
about  the  gravity  axis  parallel  to  the  flanges.  This  is  the. 
value  to  be  used  when  the  I  section  is  used  as  a  beam.  Col- 
umn 8  gives  the  moment  of  inertia  about  the  gravity  axis 
parallel  to  the  web.  The  use  of  this  value  is  discussed  in 
the  chapter  on  columns.  Column  7  gives  the  section  modulus  t 

which,  as  stated  before,  is  equal  to  the  value  of  — .     In  design- 

c 

ing  steel  beams  to  carry  uniformly  distributed  loads  the 
safe  working  stress  may  be  taken  as  16,000  Ibs.  per  sq.  in. 
Consider  the  case  of  a  steel  I  beam  which  has  a  span  of 


DESIGN  OF  BEAMS 


79 


30  ft.  and  is  to  carry  a  uniformly  distributed  load  of  30,000 
Ibs.     In  this  case  the  maximum  bending  moment  equals 


WL     30000X30X12 


=  1,350,000  in-lbs., 


and  5=16,000,  hence 

I 


1350000 = 

16000  = 


FIG.  55. 

By  reference  to  column  7,  Table  XII,  it  is  found  that  the 
18-in.  beam  weighing  55  Ibs.  per  ft.  comes  the  nearest  to 

the  desired  value  of  -,  and  hence  the  18-in.  beam  would  be 
c 

used  in  this  case. 

For  beams  subject  to  shock  a  fiber  stress  less  than  16,000 
must  be  used. 

PROB.  82.  A  floor  is  to  carry  a  load  of  250  Ibs.  per  sq.  ft. 
and  is  supported  by  steel  I  beams  having  a  span  of  20  ft.  The 
beams  are  spaced  4  ft.  center  to  center.  Using  S— 16,000,  find 
the  size  I  beam  to  be  used. 

PROB.  83.  Find  the  safe  uniformly  distributed  load  that  can 
be  carried  by  a  heavy  24-in.  I  beam  having  a  span  of  40  ft. 


80 


STRENGTH  OF  MATERIALS 


PROB.  84.  A  steel  cantilever  beam  has  a  span  of  10  ft.,  and 
carries  a  concentrated  load  at  the  end  of  the  beam  of  10,000  Ibs. 
Find  size  of  steel  I  beam  to  be  used. 


TABLE  XII 


1 

Depth 
of 
Beam. 

2 

Weight 
per  Foot. 

3 

Area  of 
Section. 

4 

Thick- 
ness of 
Web. 

Width 
of 
Flange 

6 

Ig  Axis 
1-1. 

7 

Section 
Modulus. 

Ig  Axis 
2-2. 

9 
Radius 
of  Gy- 
ration 
Axis 
2-2 

d                w                 A 
Inches.       Pounds.        Sq.  In. 

t 

Inches. 

b 
Inches. 

I 

S 

/ 

r 
Inches. 

3 

5.50 
7.50 

1.63 
2.21 

.17 
.36 

2.33 
2.52 

2.5 
2.9 

1.7 
1.9 

.46 
.60 

.53 

.52 

4 

7.50 
10.50 

2.21 

3  09 

.19 
.41 

2.66 

2.88 

6.0 
7.1 

3.0 
3.6 

.77 
1.01 

.59 
.57 

5 

9.75 
14.75 

2.87 
4.34 

.21 
.50 

3.00 
3.29 

12.1 
15.1 

4.8 
6.1 

1.23 
1.70 

.65 
.63 

6 

12.25       3.61 
17.25       5.07 

.23 

.47 

3  33 

3.57 

21.8       7.3 
26.2       8.7 

1.85 
2.36 

.72 
.68 

7 

15.00       4.42 
20.00       5.88 

.25 
.46 

3.66 
3.87 

36.2      10.4 
42.2      12.1 

2.67 
3.24* 

.78 
.74 

8 

18.00  i     5.33 
25.25       7.43 

.27 
.53 

4.00 
4.26 

56.9;     14.2 
68.0      17.0 

3.78 
4.71 

.84 
.80 

9 

21.00 
35.00 

6.31 
10.29 

.29 
.73 

4.33 

4.77 

84.9      18.9 
111.8      24.8 

5.16 
7.31 

.90 

.84 

10 

25.00 
40.00 

7.37 
11.76 

.31 
.75 

4.66 
5.10 

122.1 
158.7 

24.4 
31.7 

6.89 
9.50 

.97 
.90 

12 

31.50 
40.00 

9.26 
11.76 

.35 
.56 

5.00 
5.21 

215.8      36.0 
245.9      41.0 

9.50 
10.95 

1.01 
.96 

15 

42.00 
60.00 

12.48 
17.65 

.41 

.75 

5.50 
5.84 

441.8     58.9 
538.6      71.8 

14.62 
18.17 

1.08 
1.01 

18 

55.00 
70.00 

15.93 
20.59 

.46 

.72 

6.00 
6.26 

795.6 
921.2 

88.4 
102.4 

21.19 
24.62 

1.15 
1.09 

20 

65.00 
75.00 

19.08 
22.06 

.50 
.65 

6.25 
6.40 

1169.5 

1268.8 

117.0 
126.9 

27.86 
30.25 

1.21 
1.17 

24 

80.00 
100.00 

23.32 
29.41 

.50 
.75 

7.00 
7.25 

2087.2 
2379.6 

173.2 
198.3 

42.86 
48.55 

1.36 

1.28 

DESIGN  OF  BEAMS 


81 


ART  28.     CAST-IRON  BEAMS 

Owing  to  its  low  tensile  strength  and  low  ductility  cast 
iron  is  not  suited  for  beams  used  in  building  construction. 
It  is,  however,  used  extensively  in  the  construction  of  ma- 
chines where  heavy  bases,  frames,  or  bedplates  are  required. 
Cast  iron  is  readily  formed  into  any  desired  shape  by  casting 
and  is,  therefore,  the  material  best  adapted  to  the  purposes 
just  stated.  There  are  many  cases  where  these  machine 
parts  take  the  nature  of  a  beam,  thus  putting  the  member 
in  tension.  Grate  bars  for  furnaces  are  made  of  cast  iron, 
and  these  must  be  designed  as  beams. 

A  very  common  illustration  of  a  cast-iron  beam  is  in  the 
case  of  gear  teeth.  The  gear  blank 
is  cast  and  the  teeth  are  usually  cut, 
although  on  the  large  gears  such  as 
are  used  in  rolling  mills,  etc.,  the 
teeth  are  cast.  Fig.  56  shows  a  com- 
mon form  of  tooth.  Let  L  equal 
the  working  depth  of  the  tooth  in 
inches  and  assume  that  the  load  W 
is  concentrated  on  the  end  of  the 
tooth.  The  maximum  bending 


or 

moment  equals  WL,  and  the  resisting  moment  =  — . 

c 


Let 


F  equal  the  face  of  the  tooth  in  inches.     The  dangerous  sec- 
tion is  at  A-B  and  the  resisting  moment  equals 


hence 


WL  = 


SFh2 
6  "' 

SFh2 


or 


6 

6T7L 

Sh2  '' 


(30) 


82 


STRENGTH  OF  MATERIALS 


but  L  and  h  are  functions  of  the  circular  pitch  of  the  teeth 
so  that  the  equation  for  F  is  written 


F  = 


. 

Spy' 


(31) 


where  y  equals  a  constant,  and  p  equals  the  circular  pitch  in 
inches.  Mr.  Wilfred  Lewis  has  suggested  proper  values  of 
y  and  S  for  cast-iron  teeth.  The  student  is  referred  to  Kent's 
Handbook  for  further  data  on  gear-teeth  design. 

EXAMPLE.  In  Fig.  56,  h  =  .87  in.,  L=  1  in.,  TF  =  2400  Ibs., 
and  5  =  3000;  find  the  face  of  tooth.  Here  ft  =  .87;  L=l  in. 
and  TF  =  3600.  Substituting  these  values  in  Equation  (30) 
gives 

QWL    6X2400X1  . 


3056x37* 


The  value  of  S  depends  upon  the  speed  of  the  pitch  line  of  the 
gear,  as  well  as  upon  the  material  used. 

Bed  plates  of  lathes  and  planers  are  illustrations  of  cast- 
iron  beams.  No  set  rules  can  be  given  for  the  proportions 
to  be  used  in  cast-iron  beams,  but  as  a  general  rule  it  is  wise 
to  so  distribute  the  metal  as  to  secure  the  maximum  section 
modulus  with  the  minimum  weight  of  metal. 

PROB.  85.  A  cast-iron  gear  wheel  of  24  ins.  pitch  diameter  trans- 
mits 30  horse-power  at  200  R.P.M.  The  teeth  are  2.5  diametral 
pitch.  Find  unit  working  stress  if  the  gear  teeth  are  5  ins.  wide. 


FIG.  57. 

PROB.  86.  A  cast-iron  grate  bar  is  of  the  form  shown  in  Fig.  57. 
If  the  span  is  4  ft.  find  the  safe  uniformly  distributed  load  that 
the  bar  can  carry,  using  £  =  8000  Ibs.  per  sq.  in. 


DESIGN  OF  BEAMS  83 

PROB.  87.  Fig.  58  gives  the  cross-section  of  the  bed  of  a  lathe. 
The  span  is  6  feet.  Using  $  =  8000  Ibs.  per  sq.  in.,  find  safe  uni- 
formly distributed  load  the  bed  plate  can  carry. 


FIG.  58. 


ART.  29.    BEAMS  OF  UNIFORM  STRENGTH 

There  are  many  cases  where  a  beam  of  uniform  strength 
is  desirable;  for  example,  in  the  case  of  gear  teeth,  but  it  is 
not  always  profitable  to  secure  the  beam  of  uniform  strength 
at  the  sacrifice  of  other  important  details.  In  all  previous 
examples  the  fiber  stress  has  been  investigated  only  at  the 
dangerous  section  of  the  beam.  For  a  beam  to  be  of  uniform 
strength  it  is  necessary  that  the  fiber  stress  be  the  same  at  all 
sections  of  the  beam.  The  contour  of  the  beam  will  depend 
upon  the  nature  of  loading.  Several  cases  will  be  considered. 

CASE  I.  Simple  beam  of  uniform  strength  concentrated 
load  at  the  center  (see  Fig.  59).  Let  di=the  depth  of  the 
beam  at  the  center  of  the  span  and  6  =  the  breadth.  Let 
d  =  ihe  depth  at  any  section  distant  x  feet  from  the  left  end. 

3  PL 
From  Equation  (24)  S  =  •=  7-7-2-     At  ^ne  section  A -B  the  bend- 

p 

ing  moment  equals  ^x  and  the  resisting  moment  equals 


84  STRENGTH  OF  MATERIALS 

therefore, 


or 


2        6   ' 
3Px 


FIG.  59. 

Now  for  the  beam  to  be  of  uniform  strength,  the  value  of  S 
at  A-B  must  equal  the  value  of  S  at  the  center  of  the  span,  or 


3  PLJ&Px 

2bdi2     bd2' 


or 


_= 

2di2~d2' 


Therefore 


l2x 


(32) 


CASE  II.  Simple  beam.  Uniformly  distributed  load. 
See  Fig.  60.  As  before,  let  d\  equal  the  depth  at  the  center 
and  d  equal  the  depth  at  the  section  A-B  distant  x  feet  from 

the  left  end.     Then,  for  the  center  section  S=  -r  rj^,  (see 

4  OCt  i 


DESIGN  OF  BEAMS 


85 


Equation  (26))  at  the  section  A-B  the  bending  moment  equals 


wL 


wx 


Sbd2 


.  ,.  T  . 

—  x  —  ^—  ,  and  the  resisting  moment  equals  —  ^—  ;  hence 
22  o 


or 


bd2 


ooooooooooooooooooon 


q 


FIG.  60. 


Equating  these  values  of  S  gives 


3wL2==3w(Lx-x2) 
~ 


bd2 


or 


or 


L2  =Lx-x2 
di2~      d2    ' 


d  = 


2diVx(L-x) 


(33) 


CASE  III.  Cantilever  beam.  Concentrated  load  at 
end.  See  Fig.  61.  Let  d\  equal  the  depth  at  the  wall  and 
d  equal  the  depth  at  any  section  x  feet  from  the  left  end. 


86 


STRENGTH  OF  MATERIALS 


QPL 
Then  at  the  wall  section  S  =  T      and  at  section  A-B  the  bend 


ing  moment  equals  Px  the  resisting  moment  equals 
or 


hence 


QPx 


FIG.  61. 


Sbd2 
6    ' 


Equating  these  values  of  S  gives 


6PL_  6Px 
~~ 


or 


(34) 


CASE  IV.     Cantilever  beam.        Uniformly   distributed 
load.     See  Fig.  62.     At  the  wall  section  S=  ,  ,  2  ,  and  at  the 


DESIGN  OF  BEAMS 


87 


section  A-B  the  bending  moment  equals  wx2,  the  resisting 

Sbd2 
moment  equals  — - - ;  hence 


QQQOQOQQOOQOQOOOOOOOOOOO 


FIG.  62, 
Equating  these  values  of  S  gives 

S 


or 


therefore 


3wx2 

bdi2  ~  bd2  ' 

'T2' 
xdi 


(35) 


PROB.  88.  A  simple  wooden  beam  6  ins.  wide  carries  a 
uniformly  distributed  load  of  5400  Ibs.  Design  this  beam  so  to 
be  of  uniform  strength,  assuming  S  =  1200  and  the  span  to  be  12  ft. 


88  STRENGTH  OF  MATERIALS 

ART.  30.     MODULUS  OF  RUPTURE 

Equation  (16)  in  Art.  21  was  derived  on  the  assumption 
that  the  fiber  stress  varies  directly  with  the  given  element 
of  area  from  the  neutral  axis.  This  statement  is  true  so  long 
as  the  unit  stress  does  not  exceed  the  elastic  limit  of  the 
material.  Beyond  the  elastic  limit  the  stress  does  not  vary 
directly  with  the  distance  of  the  element  from  the  neutral 
axis.  Within  the  elastic  limit  the  value  of  S  will  agree  with 
the  tensile  or  compressive  strength  of  the  material. 

In  testing  materials  under  transverse  tests  it  is  customary 

R£c 
to  figure  the  breaking  stress  from  the  formula  S  =  -j-,  and  to 

call  the  value  thus  found  the  "  modulus  of  rupture  "  of  the 
material.  This  modulus  is  neither  equal  to  the  tensile  nor 
compressive  strength  of  the  material  and  is  not  to  be  con- 
fused with  these  values.  For  example,  the  tensile  strength 
of  cast  iron  is  about  20,000  Ibs.  per  sq.  in.  and  the  com- 
pressive strength  of  cast  iron  if  90,000  Ibs.  per  sq.  in.  The 
"  modulus  of  rupture  "  of  cast  iron  as  determined  by^  experi- 
ment average  35,000  Ibs.  per  sq.  in. 

In  the  case  of  wrought  iron  and  mild  steel  there  is  no 
fixed  modulus  of  rupture,  as  these  materials  will  continue 
to  bend  under  the  action  of  a  transverse  load.  In  such  cases 
the  load  carried  is  generally  fixed  by  the  allowable  deflection. 
For  beams  a  safe  working  stress  of  16,000  can  be  used  for 
mild  steel. 

In  conducting  transverse  tests  of  wood  it  is  essential 
that  the  span  be  at  least  ten  times  the  depth  of  the  beam. 
The  ends  of  the  specimen  should  rest  on  bearing  plates  rather 
than  directly  on  the  knife-edges  of  the  testing  machine.  For 
any  beam  of  rectangular  cross-section  tested  under  a  con- 
centrated load  of  P  pounds  at  the  center  of  a  span  of  L  inches, 
the  modulus  of  rupture  is  found  from  the  formula 


DESIGN  OF  BEAMS 


89 


where  P  equals  the  load  required  to  cause  rupture  of  the 
specimen.  Table  XIII  gives  the  modulus  of  rupture  of 
various  materials. 

TABLE  XIII 
MODULUS  OF  RUPTURE 


Material. 

Bending  Pounds 
Per  Sq.  In. 

Torsion  Pounds 
Per  Sq.  In. 

Timber.                     .          

9,000 

Cast  iron. 

35000 

Wrought  iron    

40,000 

Copper  cast                      

30,000 

Granite 

1  900 

EXAMPLE.  A  2X4-in.  yellow  pine  beam  of  42-in.  span 
broke  under  a  concentrated  load  of  5000  Ibe.  at  the  center 
of  the  span.  Find  the  modulus  of  rupture.  Here  P  =  5000, 
6  =  2  ins.,  h  =  4  ins.,  L  =  42  ins.,  hence 


3  PL    3 

S=2  T/T  2 


5000X42 


In  the  case  of  a  section  which  is  not  symmetrical  with 
respect  to  the  gravity  axis,  it  is  usual  to  assume  that  Equa- 
tion (16)  is  true,  and  figure  values  of  S  for  both  top  and  bot- 
tom fibers.  For  example,  in  the  case  of  the  cast-iron  T 
section  shown  in  Fig.  63.  This  section  was  tested  as  a  beam 
of  12-in.  span,  and  broke  under  a  load  of  2400  lbs.  The 
bottom  fibers  were  in  tension  and  the  top  fibers  were  in 
compression.  The  maximum  fiber  stress  in  tension  is  found 

from  the  equation  S  =  —^  ,  using  for  c  the  distance  from  the 

neutral  axis  to  the  base  of  the  beam,  which  in  this  case  is 
46  in.     The  bending  moment 

,,     PL     2400X12     7___ 
M  =  •—  =  —          -  =  7200  in  .-lbs. 
4  4 


90 


STRENGTH  OF  MATERIALS 


The  moment  of  inertia  relative  to  the  gravity  axis  is  approxi- 
mately .08.  Substituting  these  values  in  the  above  equation 
gives 

„     7200  X.  46 


— 


=  41,400  Ibs  per  sq.  in. 

,25'k- 


PROB.  89.  A  piece  of  white  pine  2x4  ins.  cross-section  and 
42  ins.  span  broke  under  a  transverse  concentrated  load  .of  5000 
pounds.  Find  the  modulus  of  rupture. 

PROB.  90.  A  cast-iron  T  section  of  form  shown  in  Fig.  63, 
having  the  following  dimensions,  length  flange  =1.25  ins.,  depth  of 
section  =  1.25  ins.,  thickness  of  we  band  flange  =  .30  ins.,  span  12  ins. 
broke  under  a  load  of  2200  Ibs.  Find  the  maximum  fiber  stress  in 
tension  and  compression.  Load  applied  on  top  of  web,  flange 
horizontal. 

ART.  31.     PRACTICAL  PROBLEMS 

PROB.  91.  A  floor  carrying  a  load  of  100  Ibs.  per  sq.  ft.,  is 
supported  by  3X8-in.  yellow-pine  beams  of  12-ft.  span.  Find 
proper  spacing  of  beams. 

PROB.  92.  The  beams  in  Prob.  91  are  supported  by  wooden 
girders  of  24-ft.  span.  The  depth  of  the  girders  is  twice  the  breadth. 
Find  the  dimensions  of  the  girders. 

PROB.  93.    A  floor  is  supported  by  steel  I  beams,  of  20-ft.  span 


DESIGN  OF  BEAMS 


91 


and  6-ft.  centers.    The  floor  sustains  a  load  of  200  Ibs.  per  sq.  ft. 
Find  size  I  beams  required. 

PROB.  94.    A  yellow-pine  beam  is  loaded  as  shown  in  Fig.  64. 
Find  the  maximum  fiber  stress  and  the  factor  of  safety. 


2500 


-20- 


1 8X12" 


FIG.  64. 


PROB.  95.  Fig.  65  shows  the  arrangement  of  a  pin  connection 
in  a  bridge  truss.  Find  diameter  of  the  pin,  using  £  =  8000  Ibs, 
per  sq.  in. 


—4  + 


rti 


r 


r 


W.I.  Pin 


i  P 
FIG.  65. 


PROB.  96.  A  12-in.  steel  I  beam  of  30-ft.  span  supports  a  load 
of  500  Ibs.  per  lin.  ft.  Is  the  beam  safe? 

PROB.  97.  A  traveling  crane  consists  of  two  heavy  18-in.  I 
beams  of  40-ft.  span.  Neglecting  the  weight  of  the  trolley  and 


92 


STRENGTH  OF  MATERIALS 


hoist,  find  maximum  load  that  can  be  raised  so  that  the  safe  fiber 
stress  shall  not  exceed  12,000  Ibs.  per  sq.  in. 

PROB.  98.  A  cast-iron  beam  1|X2|  ins.  in  section  and  30  ins. 
span  broke  under  a  concentrated  load  of  7000  Ibs.  Find  the  mod- 
ulus of  rupture. 


FIG.  66. 

PROB.  99.  Fig.  66  shows  the  crank  pin  of  a  side  crank  engine. 
Find  the  proper  diameter  of  steel  pin  when  the  load  P  =  24,000  Ibs, 
Find  the  factor  of  safety  in  shear. 

PROB.  100.  Compare  the  safe  loads  that  can  be  carried  by  a 
yellow-pine  beam  8X10  ins.  in  cross-section  and  20-ft.  span,  and  a 
light  10-in.  I  beam  of  30-ft.  span. 


CHAPTER  VI 
DEFORMATIONS 

ART.  33.    MODULUS  OF  ELASTICITY 

THE  term  deformation  is  used  to  designate  the  change  of 
form  of  any  part  of  a  machine  or  structure  when  subjected 
to  an  external  load.  As  stated  previously  in  the  case  of 
tension  the  deformation  becomes  an  elongation,  in  the  case 
of  compression  the  deformation  is  a  shortening  of  the  part; 
in  the  case  of  bending  the  deformation  takes  the  form  of  a 
deflection  of  the  neutral  axis  of  the  part;  and  in  the  case  of 
torsion  the  deformation  becomes  a  twisting  action  of  the 
external  fibers  of  the  part.  T  he  relation  between  the  external 
load  applied  and  the  resulting  deformation  is  not  always  a 
simple  one  to  determine,  but  in  general  this  relation  can  be 
expressed  as  a  ratio,  which  is  referred  to  as  the  "  modulus  of 
elasticity." 

The  modulus  of  elasticity  is  equal  to  the  ratio  of  the  unit 
stress  to  the  unit  deformation  at  the  elastic  limit  of  the  mate- 
rial. So  long  as  the  load  does  not  exceed  the  elastic  limit, 
this  ratio  is  assumed  to  be  constant.  However,  under  actual 
test  the  ratio  is  found  to  vary  for  the  lighter  loads.  Let  E 
equal  the  modulus  of  elasticity,  S  equal  the  unit  stress  in 
pounds  per  square  inch,  and  d  equal  the  unit  deformation, 
then 

E  =  ^ (36) 

or 

d  =  |.      .......     (37) 

93 


94  STRENGTH  OF  MATERIALS 

In  the  case  of  a  tension  test  let  P  equal  the  load  at  the  elastic 
limit,  L  equal  the  length  in  inches  of  the  part  of  the  specimen 
under  test,  e  equal  the  total  elongation  at  the  elastic  limit 
and  A  equal  the  original  cross-section  of  the  specimen;  then 


and 


or 


For  example,  a  specimen  of  the  form  shown  in  Fig.  12  gave 
the  following  results  in  a  tension  test;  load  at  elastic  limit 
was  7000  Ibs.,  elongation  at  elastic  limit  .0024  in. 

The  unit  stress  at  the  elastic  limit  equals  -j  =  —  —^  =  35,800. 

A.     .  iyb 

and 

0024 
the  unit  deformation  equals      2     =  0012  in.; 

hence 

.<?    s^snn 

,  =  30,000,000  (approx.). 


The  modulus  of  elasticity  is  an  index  of  the  stiffness  of  the 
material,  so  that  if  any  machine  part  is  under  tension  and 
the  unit  stress  does  not  exceed  the  elastic  limit  of  the  mate- 
rial, the  probable  elongation  of  the  part  can  be  figured  if  the 
modulus  of  elasticity  is  known. 

For  example,  consider  the  case  of  a  steel  tension  member 
to  be  1|X4  ins.  cross-section  and  10  ft.  long,  and  carrying  a 
load  of  72,000  Ibs.  The  unit  stress  in  this  case  equals 

72000 
— « —  =  12,000  Ibs.  per  sq.  in.     The  total  elongation  which 


DEFORMATIONS 


95 


will  probably  occur  may  be  found  by  arranging  Equation 
(38)  in  the  form 

(39) 


thus 


e  — 


AE 
72000X120 


6X30000000 


~  =.048  in. 


In  the  design  of  any  structure  proper  allowance  must  be 
made  for  the  increase  in  length  of  the  various  members  when 
subjected  to  a  tensile  stress. 

In  like  manner  the  probable  shortening  can  be  deter- 
mined in  the  case  of  a  compression  test.  In  this  case  the 
value  of  e  in  Equation  (38)  is  to  be  taken  as  the  decrease  in 
length  of  the  specimen,  and  the  other  terms  are  the  same  as  in 
tension.  However,  this  equation  holds  true  only  when  the 
specimen  is  under  pure  compression,  and  must  not  be  con- 
fused with  the  case  where  the  specimen  acts  as  a  column. 
(See  Chapter  VIII.) 

The  values  of  the  modulus  of  elasticity  are  given  in  Table 
XIV.  It  will  be  noted  that  the  modulus  is  about  the  same 
in  tension  and  compression.  In  the  case  of  torsion  the  term 
modulus  of  rigidity  is  sometimes  used. 

TABLE  XIV 

MODULUS  OP  ELASTICITY. 


Material. 

Tension. 

Compression. 

Shear. 

White  pine 

1  130  000 

1,130,000 

Yellow  pine  

1,500,000 

1,500,000 

White  oak  
Cast  iron. 

1,150,000 
12,000,000 

1,150,000 
12,000,000 

Wrought  iron 

28  000  000 

28  000  000 

Copper  wire  
Steel  (plate)          . 

18,000,000 
29,000,000 

18,000,000 
29,000,000 

Steel  (machine) 

30  000,000 

30  000  000 

96 


STRENGTH  OF  MATERIALS 


PROB.  101.  A  steel  specimen  0.8  in.  in  diameter  and  8  ins. 
long  was  subjected  to  a  load  of  18,000  Ibs.  with  a  resulting  elonga- 
tion of  .0096.  Find  the  modulus  of  elasticity.  * 

PROB.  102.  How  much  will  a  wrought-iron  bar  1X2  in.  and 
6  ins.  long  shorten  under  a  load  of  50,000  Ibs.? 

ART.  33.     DEFLECTION  OF  BEAMS 

The  relation  between  the  unit  stress  and  the  unit  deforma- 
tion in  the  case  of  beams  is  not  as  easily  determined  as  in  the 
case  of  tension  and  compression.  When  a  beam  is  subjected 
to  a  bending  moment  the  lower  fibers  are  put  in  tension  and 
the  bottom  part  of  the  beam  is  elongated,  while  the  top  part 
is  shortened.  The  neutral  axis  remains  the  same  length. 
Thus,  in  Fig.  67  ox  represents  the  position  of  the  neutral 


o     \ 


\E 


FIG.  67. 


axis  when  the  external  bending  moment  is  zero  and  the  line 
oMNx  the  position  of  the  neutral  axis  under  the  action  of 
the  external  bending  moment.  At  points  M  and  N  con- 
struct the  normals  OA  and  OE.  These  normals  intersect 
at  some  point  0,  which  is  the  center  of  curvature  for  the 
line  MN.  The  distance  ab  represents  the  deformation 
for  the  element  of  length  represented  by  MN  or  I.  The 

deformation  per  unit  of  length  equals  -=-,  hence  the  modulus 

I 

of  elasticity  equals  the  unit  stress  divided  by  the  unit  deforma- 


DEFORMATIONS 


97 


07  07- 

tion,  or  E  =  —r,  or  a  b=-^.     If  I  is  taken  as  a  very  short 

CLO  Ei 

distance,  it  will  be  practically  a  straight  line,  and  then  the 
triangle   OMN   and   the   triangle   Nab   are   similar,    hence 

R     c    ,         ,     SL    .,       ,       R     c-E        R    E        , 

T  =  -r,  but  a&  =  — .  therefore  -7  =  -^7-  or  —  =  -^.     The  unit 

I      ao  Ei  i       oL/          c      o 

Afc 

stress  S  =  -,-- ',  substituting  this  value  of  S  in  the  above  equa- 
tion gives 


c     Me1 


or 


R 


(40) 


Equation  (40)  is  referred  to  as  the  equation  of  the  elastic 
curve.  The  value  of  R  will  depend  upon  the  nature  of  loading 
of  the  beam.  Its  value  can  be  computed  only  by  the  aid  of 
the  calculus.  It  can  be  seen,  however,  that  the  deflection  d 
of  the  beam  will  depend  upon  the  radius  of  curvature  R  of 
the  elastic  curve. 


FIG.  G8. 

In  the  case  of  a  cantilever  beam  with  a  load  of  P  pounds 
at  the  end  of  a  beam  of  L  inches  length,  the  maximum  deflec- 
tion will  be  directly  under  the  load  P  as  shown  in  Fig.  68. 
The  value  of  this  deflection  is  given  by  the  formula 


(41) 


3EF 


98  STRENGTH  OF  MATERIALS 

where  P  =  the  load  in  pounds; 
L  =  the  length  in  inches; 
#=the  modulus  of  elasticity; 
7  =  the  moment  of  inertia  of  the  section,  and 
d  =  the  maximum  deflection  in  inches. 

As  an  application  of  the  formula  consider  the  case  of  a 
6X9  ins.  yellow-pine  cantilever  beam,  which  extends  6  ft. 
beyond  the  support  and  carries  a  concentrated  load  of  1350 
Ibs.  The  maximum  deflection  in  this  case  equals 

PL3 
3EF 

Where  P=  1350; 

1,3=723=373,248; 

#=1,500,000; 
and 

/  =  6d3  =  6X729 

hence 


1350X373248 
3  X 1500000  X  364 . 5  m> 


Equation  (41)  can  also  be  written  in  terms  of  the  maxi- 

Mc 
mum  unit  fiber  stress.  .  From   Equation  (16)  S=-j-,   and 

in  the  case  of  a  cantilever  beam  with  concentrated  load  at  the 
end  M  =  PL,  hence 

c_Mc_PLc 

:T=~/~ 

or 

M 
Lc 

Substituting  this  value  in  Equation  (41)  gives 
57    If       SL* 


DEFORMATIONS  09 

From  this  equation  it  is  seen  that  for  a  given  fiber  stress 
the  deflection  of  a  beam  varies  directly  as  the  square  of  the 
length. 

For  a  cantilever  beam  with  a  uniformly  distributed  load 
of  W  pounds  the  deflection  is  given  by  the  formula, 


Here  again,  the  deflection  may  be  expressed  in  terms  of 

the  unit  stress  for 

Me 

=  7~' 

but 

M-^- 

2   ' 

hence 

TFL  c     WLc 
2  'I      21  ' 
or 


. 

Lc 

Substituting  this  value  in  Equation  (43)  gives 


_ 
~8LcEI~4Ec' 

For  a  simple  beam  with  a  concentrated  load  of  P  pounds 
at  the  center  of  the  span  the  maximum  deflection  is  given  by 
the  equation 


or  in  a  similar  manner  to  the  above  it  can  be  shown  that 
,     SL* 


_, 
(46) 


100  STRENGTH  OF  MATERIALS 

For  a  simple  beam  with  a  uniformly  distributed  load  of  W 
pounds  the  deflection  is  given  by  the  equation 


or 


PROB.  103.  Find  the  maximum  deflection  of  a  12-in.  heavy 
steel  I  beam  of  25-ft.  span  which  carries  a  uniformly  distributed 
load  of  10,000  Ibs. 

PROB.  104.  A  steel  shaft  6  ins.  diameter  rests  on  two  bearings 
which  are  6  ft.  apart.  The  shaft  carries  a  flywheel  weighing  2400 
Ibs.  located  at  the  center  of  the  span.  Find  the  deflection  due  to 
the  weight  of  the  flywheel. 

ART.  34.     PROBLEMS 

PROB.  105.  Assume  the  diameter  of  the  crank  pin  in  Fig.  66 
equals  4  ins.  The  maximum  allowable  deflection  at  the  end  of  the 
pin  is  .01  in.  Find  the  load  P  required  to  give  this  deflection. 

PROB.  106.  A  cast-iron  water  pipe  24  ins.  in  diameter  rests 
on  supports  which  are  located  20  ft.  apart.  Find  the  deflection  of 
the  pipe  when  filled  with  water.  Thickness  of  the  pipe  equals 
If  ins. 

PROB.  107.  A  15-in.  heavy  steel  I  beam  of  30-ft.  span  carries 
a  uniformly  distributed  load.  Find  the  value  of  this  load  so  that 
the  maximum  deflection  shall  not  exceed  sino  of  the  span. 

PROB.  108.  A  floor  is  supported  by  3X8  wooden  beams  spaced 
18  ins.  apart.  Find  the  load  the  floor  can  carry  in  pounds  per  square 
foot,  so  that  the  deflection  of  the  ceiling  supported  under  the  floor 
will  not  exceed  the  allowable  value  of  3^  of  the  span. 

PROB.  109.  Two  beams  of  uniform  cross-section  and  having 
span  of  10  ft.  and  20  ft.  respectively  carry  concentrated  loads 
producing  equal  fiber  stresses  in  both  beams.  Find  the  relative 
deflection  of  the  two  beams. 

PROB.  110.  In  a  transverse  test  of  a  simple  wood  beam  of 
42-in.  span  the  load  at  the  elastic  limit  was  2400  Ibs.,  and  the  cor- 


DEFORMATIONS,  •  \         ;  > ,'  V  '  •  ] ,' ;  '. ',  1Q1 ' 

responding  deflection  0.154  in.  Find  the  modulus  of  elasticity. 
Beam  was  3  ins.  in  breadth  and  4  ins.  in  depth.  Load  was  applied 
at  the  center  of  the  span. 

PROB.  111.  Determine  the  spacing  of  light  10-in.  I  beams 
which  earn7  a  uniform  floor  load  of  100  Ibs.  per  sq.  ft.,  so  that  the 
maximum  deflection  shall  not  exceed  -3-^  of  the  span.  The  beams 
are  18-ft.  span. 

PROB.  112.  In  Fig.  48  find  the  maximum  deflection  when  the 
load  P  equals  3000  Ibs. 

PROB.  113.  A  box  girder  of  the  form  shown  in  Fig.  49  has  a 
span  30  ft.  Find  the  uniformly  distributed  load  required  to  pro- 
duce a  maximum  deflection  of  0.75  in.  at  the  center  of  the  span, 
Plates  are  14  ins.  wide. 

PROB.  114.  In  a  transverse  test  of  a  cast-iron  beam  IX 1  in. 
in  section  and  12  ins.  span  the  load  at  the  elastic  limit  was  2800 
Ibs., 'and  the  deflection  0.08  in.  Find  the  modulus  of  elasticity  of 
cast  iron.  Load  applied  at  the  center  of  the  span. 


CHAPTER  VII 
SHAFTING 

ART.  35.    TWISTING  MOMENT 

WHEN  an  external  load  is  applied  tangential  to  the  cir- 
cumference of  a  pulley  there  is  produced  a  tendency  to  dis- 
tort the  shaft  on  which  the  pulley  is  mounted;  that  is,  the 
neutral  axis  of  the  shaft  will  not  be  affected,  but  the  outer 
fibers  of  the  shaft  will  tend  to  take  the  form  of  spirals. 
Under  such  a  condition  the  shaft  is  said  to  be  under  torsion, 
and  the  stress  produced  in  the  shaft  is  one  of  shear.  The 
product  of  the  tangential  force  times  the  radius  of  the  pulley 
is  called  the  twisting  moment,  which  moment  may  be  nega- 
tive or  positive. 

The  amount  of  distortion  or  twist  of  the  outer  fiber  of 
the  shaft  is  proportional  to  the  external  load  applied,  pro- 
vided the  elastic  limit  of  the  material  is  not  exceeded. 


FIG..  69. 

The  twisting  moment  at  various  sections  of  a  shaft  is  not 
necessarily  the  same.  For  example,  in  Fig.  69  let  C,  Z>,  and 
E  be  pulleys  mounted  upon  a  shaft  to  which  power  is  delivered 

102 


SHAFTING 


103 


by  the  coupling  A .  Assume  that  belts  are  placed  on  the  pul- 
leys and  that  the  belt  pulls  are  200,  150,  and  100  Ibs.  on  the 
pulleys  C,  D,  and  E,  respectively,  and  that  the  belts  are 
traveling  in  the  same  direction.  The  twisting  moment  on 
the  section  between  the  pulley  E  and  the  bearing  F  will 
equal  zero;  between  the  pulleys  D  and  E  the  twisting 
moment  equals  4X100  =  400  in.-lbs.  Between  pulleys  C 
and  D  the  twisting  moment  equals  6X150+4X100=1300 
in.-lbs.  Between  pulley  C  and  the  coupling  A  the  twisting 
moment  equals  12X200+150X6+4X100  =  3700  in.-lbs. 

The  twisting  moment  on  a  shaft  is  sometimes  referred  to 
as  the  torque.  For  instance,  in  the  case  of  a  motor  shaft, 
the  magnetic  pull  of  the  field  tends  to  rotate  the  armature  of 
the  motor,  which,  in  turn,  produces  a  twisting  moment  or 
torque  on  the  shaft. 

Test  specimens  are  tested  in  torsion  by  placing  the  speci- 
men in  jaws  which  in  turn  are  placed  in  the  heads  of  a  testing 
machine.  One  of  these  heads  is  fixed  and  the  other  is  slowly 
rotated,  setting  up  a  torque  or  twist  in  the  specimen.  The 
angular  deformation  is  measured  by  the  number  of  degrees 
through  which  the  movable  head  is  turned. 


FIG.  70. 

Twisting  moments  like  bending  moments  should  be 
figured  in  inch-pounds.  In  Fig.  70,  let  PI,  PI  and  P%  repre- 
sent forces  acting  at  distances  of  pi,  P2,P3  from  the  center  of 
the  shaft.  The  force  PI  tends  to  rotate  the  shaft  in  a  clock- 


104  STRENGTH  OF  MATERIALS 

wise  direction,  and  produces  a  positive  twisting  moment. 
The  forces  Pi  and  PS  tend  to  rotate  the  shaft  counter-clock- 
wise, and  produce  negative  twisting  moments.  The  result- 
ant twisting  moment  is  the  algebraic  sum  of  the  moments  of 
all  the  forces  tending  to  twist  or  rotate  the  shaft.  In  this 
case  the  resultant  twisting  moment  equals  -\-P\p\  —  P2p2  — 
Psps,  or,  expressed  as  an  equation, 

T=2Pp (49) 

It  is  to  be  noted  that  in  figuring  twisting  moments  the 
moment  arm  is  taken  as  the  perpendicular  distance  from  the 
center  of  the  shaft  to  the  line  of  action  of  the  force. 

PROB.  115.  A  pulley  48  ins.  in  diameter  is  placed  on  the  end 
of  a  shaft.  A  belt  having  an  effective  belt  pull  of  250  Ibs.  is  placed 
on  the  pulley,  Find  the  twisting  moment  produced  in  the  shaft. 

ART.  36.     RESISTING  MOMENT  IN  TORSION 

A  twisting  moment  produces  a  shear  in  the  cross-section 
of  a  shaft  which  varies  from  0  at  the  neutral  axis  to  "a  maxi- 
mum at  the  outer  fiber  or  circumference  of  the  shaft.  Within 
the  elastic  limit  of  the  material  this  unit  shear  is  proportional 
to  the  distance  of  the  fiber  from  the  neutral  axis.  Let  Fig.  71 
represent  the  cross-section  of  a  shaft  at  which  point  the 
external  twisting  moment  equals  PXp.  Let  S  =  the  fiber 
stress  on  the  outermost  fiber  of  the  shaft  and  let  a  equal  an 
element  of  the  cross-section  of  the  shaft  located  at  a  distance 
of  z  units  from  the  neutral  axis,  which  passes  through  the 
point  0.  Assuming  that  the  elastic  limit  of  the  material  is 
not  exceeded,  the  unit  stress  on  the  element  of  area  a  equals 

S  -,  where  c  is  the  distance  from  the  neutral  axis  to  the  outer- 
c 

most  fiber.  The  total  stress  on  this  small  element  of  area 
equals  S—a.  As  the  external  twisting  moment  tends  to 

0 


SHAFTING 


105 


shear  the  shaft,   the  internal   stress    on  area  a  tends   to 
exert  an  internal  resisting  moment,  which  moment  equals 

the  stress  S--a,  times  its  moment  arm  z,  or  the  resisting 
c 

moment  of  the  small  element  of  area  equals  S--a-z  =  —  az2. 

c  c 

Now  consider  the  cross-section  of  the  shaft  to  be  made  up  of 
N  areas  of  the  form  a,  and  let  A  equal  the  cross-sectional 


p ^ 


FIG.  71. 

area  of  the  shaft.  Then  A=--aN.  The  total  resisting  mo- 
ment of  the  entire  cross-section  equals  the  sum  of  the  resisting 
moments  of  the  elementary  areas.  Let  T  equal  this  resisting 
moment.  Then 


Let  Z2  =  the  average  value  of  the  terms  zi2,  Z22,  za2,  etc. 


106  STRENGTH  OF  MATERIALS 

Then 
or 


but 

aN  =  A, 
hence 


.The  expression  AZ2  is  called  the  polar  moment  of  inertia  of 
the  section.  Let  J  equal  the  polar  moment  of  inertia.  For 
equilibrium  to  exist  the  external  twisting  moment  must  equal 
the  internal  resisting  moment,  hence 


(50) 


Equation  (50)  is  similar  to  Equation  (15),  and  as  Equation 
(15)  is  fundamental  in  the  design  of  beams,  so  Equation 
(50)  becomes  fundamental  for  the  design  of  shafts  subjected 
to  torsion  only. 

PROB.  116.  A  shaft  6  ins.  in  diameter  is  subjected  to  an  external 
twisting  moment  such  that  the  maximum  shear  on  the  outermost 
fiber  is  3000  Ibs.  per  sq.  in.  What  is  the  average  shear  on  each 
square  inch  of  cross-section? 


ART.  37.    POLAR  MOMENT  OF  INERTIA 

The  polar  moment  of  inertia  of  any  section  is  equal  to 
the  sum  of  the  products  obtained  by  multiplying  each 
element  of  area  by  the  square  of  its  distance  from  the  center 
of  the  section.  The  term  polar  is  used  because  the  distances 
are  measured  from  a  point  or  pole.  In  the  design  of  beams 
the  moment  of  inertia  was  obtained  by  multiplying  each 
element  of  area  by  the  square  of  its  distance  from  a  given 
axis.  This  moment  of  inertia  is  called  the  rectangular  mo- 


SHAFTING 


107 


ment  of  inertia.  The  polar  moment  of  inertia  of  a  section 
equals  the  sum  of  the  rectangular  moments  of  inertia  taken 
about  axes  which  pass  through  the  center  of  the  section 
and  are  perpendicular  to  each  other.  In  Fig.  72  let  a  equal 
an  element  of  area  located  at  a  distance  of  z  units  from  the 
center  of  the  surface,  and  at  a  distance  of  x  units  from  the 
axis  OF  and  a  distance  of  y  units  from  the  axis  OX.  Then 
by  definition  the  rectangular  moment  of  inertia  relative  to 
the  OY  axis  equals  ly  =  2az2.  In  like  manner  the  rectangular 
moment  of  inertia  about  the  OX  axis  equals  Ix  =  2ay2.  Now, 
by  definition  the  polar  moment  of  inertia  of  the  section 


FIG.  72. 
relative  to  the  center  0  equals  J=2az2.    By  inspection  of 

4-l-»,*"fc    •£  SOT  i -vvsx  -1-4-    -i^  y-k-»ri/-l^\-*-v4-    4-V*o4-    ^2  —  rt*2»/if2*     Tlf^TlPO 

I,.  (5D 


the  figure  it  is  evident  that  z2  =  x2  =  y2', 


or    J  = 


Equation  (51)  gives  a  simple  method  for  finding  the  polar 
moment  of  inertia  when  the  rectangular  moment  is  known. 
Thus,  in  the  case  of  a  circle  of  diameter  d\  the  rectangular 
moment  of  inertia  equals 


also 


108 
hence 


STRENGTH  OF  MATERIALS 


.     .     (52) 


For  sections  which  are  symmetrical  with  respect  to  the  x  and 
y  axes,  the  polar  moment  of  inertia  equals  twice  the  rect- 
angular moment. 

For  the  hollow  circular  section  shown  in  Fig.  73  the  polar 
moment  of  inertia, 


(53) 


FIG.  73. 


PROB.  117.  Find  the  polar  moment  of  inertia  of  a  square  in 
terms  of  the  length  of  one  side.  Also  find  polar  moment  for  a 
hollow  square. 

PROB.  118.  Find  the  polar  moment  of  inertia  of  a  hollow  steel 
shaft  12  ins.  in  outside  diameter  and  9  ins.  inside  diameter. 

PROB.  119.  Find  the  diameter  of  a  solid  shaft  whose  cross- 
section  has  a  polar  moment  of  inertia  equal  to  that  of  the  shaft 
in  Prob.  118.  Compare  the  relative  weights  of  the  shafts. 


SHAFTING  109 


ART.  38.     SHAFTS  IN  PURE  TORSION 

When  a  shaft  is  subjected  to  torsion  only,  the  internal 
stress  in  the  shaft  is  one  of  shear,  and  the  maximum  fiber 
stress  in  shear  is  given  by  the  equation, 


where  T  equals  the  maximum  twisting  moment  in  inch- 
pounds,  c  equals  the  distance  from  the  center  of  the  shaft 
in  the  outermost  fiber  and  J  equals  the  polar  moment  of 
inertia  of  the  section.  It  is  more  convenient  to  express  the 
unit  stress  in  terms  of  the  horse-power  transmitted  and  the 
speed  of  the  shaft.  To  derive  this  relation  several  definitions 
are  necessary. 

Work  is  the  overcoming  of  an  external  resistance  through 
a  given  space.  The  unit  of  work  is  the  foot-pound,  which  is 
the  overcoming  of  a  resistance  of  1  Ib.  through  a  distance  of 
1ft. 

Power  is  the  time  rate  of  doing  work,  and  the  common 
unit  of  power  is  the  horse-power,  which  is  the  equivalent  of 
33,000  ft.-lbs.  of  work  per  minute.  For  example,  a  weight 
of  5  tons  is  hoisted  at  the  rate  of  150  ft.  per  minute.  Here 

,      5X2000X150     0_0 

the    horse-power    developed    equals        —  00      ---  =27.3. 

ooUUU 

In  Fig.  71  assume  that  a  shaft  d  inches  in  diameter  is  being 
rotated  at  the  rate  of  N  revolutions  per  minute  by  a  force  of 
P  pounds,  acting  at  a  distance  of  p  feet  from  the  center  of  the 
shaft.  The  work  done  per  revolution  equals  2irp  •  P  ft.-lbs. 
since  the  force  P  moves  through  a  distance  equal  to  the 
circumference  of  a  circle  of  radius  p.  The  work  done  per 
minute  equals  2-n-p-PN  ft.-lbs.,  hence  the  horse-power, 


~ 


33000 


110  STRENGTH  OF  MATERIALS 

or 

33000  XH 


where  Pp~equals  the  twisting  moment  in  foot-pounds.  If 
the  twisting  moment  is  expressed  in  inch-pounds,  Equation 
(56)  becomes 

12  X  33000  XH    63025# 


SJ 

but  from  Equation  (50)  Pp  =  —  , 

c 

hence 

SJ    63025ff 

—  =   ~^~~  .......     (58) 

For  a  solid  shaft  of  diameter  d  inches,  J=  Q^-  and  c  =  ^', 

•  >-  2i 

substituting  these  values  in  Equation  (58)  gives 

7T^ 

2      63025# 


N     ' 
2 

321000^ 


(59) 
or 

Sd*N 


For  example,  a  2-in.  steel  shaft  running  at  225  R.P.M.  will 
transmit  33.6  horse-power  with  a  fiber  stress  of  COOO  Ibs.  per 


sq.  n. 
For 


„    6000  X23X  225    QQ  „ 
321000 


The  value  to  be  used  for  S  will  depend  upon  the  speed  of 
rotation.  As  a  general  rule  at  the  higher  speeds  of  rotation 
a  higher  factor  of  safety  is  desirable. 


SHAFTING  111 

As  an  example  in  design  let  it  be  required  to  find  the 
diameter  of  a  solid  steel  shaft  to  transmit  100  horse-power  at 
200  R.P.M.,  using  the  safe  working  stress  as  6000  Ibs.  per 
sq.  in.  Here  #  =  100,  W  =  200,  and  S  =  6000.  Substituting 
these  values  in  Equation  (59)  gives 

6000xd3=32iooxio 


or 

«    321000X100    97 

=  6000X200  = 
hence 


/or  a  hollow  shaft  of  outside  diameter  d\t  and  inside 
diameter  d,  the  polar  moment  of  inertia 


and 


d 
c=2' 


hence  Equation  (58)  becomes 

flx^W-*)   63025g 

di  N     ' 

2 
or 


di  N 

and 


MH  (61) 


321000^!    ...... 

For  example,  consider  the  case  of  the  tail  shaft  of  a  marine 
engine.  Inside  diameter  of  shaft  is  9  ins.  and  outside  diam- 
eter is  12  ins.  R.P.M.  =  90  and  S  =  7500  Ibs.  per  sq.  in.  The 


112  STRENGTH  OF   MATERIALS 

horse-power  transmitted  is  found  by  substituting  these  values 
in  Equation  (62).     Thus, 

7500X(124-9*)X90 
321000X12 

PROB.  120.  A  line  shaft  2|  ins.  diameter  runs  at  225  R.P.M. 
What  horse-power  can  be  transmitted  with  a  fiber  stress  of  7000  Ibs. 
per  sq.  in.? 

PROB.  121.  Compare  the  horse-power  transmitted  by  solid 
and  hollow  shafts  of  equal  areas  Diameter  solid  shaft  is  8  ins. 
and  outside  diameter  of  hollow  shaft  is  12  ins.  Assume  shafts 
run  at  same  speed  and  fiber  stress  is  same  in  each  case. 


ART.  39.     SHAFTS  FOR  COMBINED  TORSION  AND  BENDING 

In  many  cases  a  shaft  is  subjected  to  both  a  bending 
moment  and  a  twisting  moment.  For  example,  the  crank 
shaft  of  an  engine  is  subjected  to  a  twisting  moment  due  to 
the  thrust  of  the  connecting  rod  on  the  crank  pin,  and  also 
to  a  bending  moment  due  to  the  weight  of  the  flywheel,  or 
the  revolving  element  of  a  generator.  In  such  cases  it  is 
necessary  to  figure  the  maximum  bending  moment  and  the 
maximum  twisting  moment  and  then  determine  the  equiva- 
lent bending  or  twisting  moment  which  will  produce  the  same 
effect  as  the  combined  action  of  the  bending  and  twisting 
moments. 

Let  M  equal  the  maximum  bending  moment  in  inch- 
pounds  and  T  equal  the  maximum  twisting  moment.  Also 
let  Te  equal  the  equivalent  twisting  moment  in  inch-pounds 
due  to  the  combined  action  of  the  bending  moment  M,  and 
the  twisting  moment  T;  and  Me  equal  the  equivalent  bend- 
ing moment  in  inch-pounds. 

Then 

M     1     , 

....     (63) 


SHAFTING  113 

and 


(64) 


S7  SJ 

From  Equation  (15)  M  =  —  and  from  Equation  (50)  T=  —  . 

c  c 

For  a  solid  shaft  these  formulae  become 

•  •    (65) 

and 

Sd3 


When  the  shaft  is  subjected  to  both  twisting  and  bending  the 
equivalent  moments  are  to  be  used,  and  the  above  equations 
become 

Me=j^,      ...  .    (67) 

or  _ 

3/10.2Me 

d=<U  -  £  -  ,  000,0.  (68) 

and 

Te=fr       ........     (69) 

or 


In  a  given  problem  the  diameter  of  shaft  is  to  be  figured 
from  Equations  (68)  and  (70),  and  the  larger  value  to  be 
taken  as  the  required  diameter  of  shaft. 

EXAMPLE.  Consider  the  case  of  a  crank  shaft  where  the 
twisting  moment  is  300,000  in.-lbs.  and  the  bending  moment 
240,000  in.-lbs.  Let  it  be  required  to  find  the  diameter  of  a 
solid  shaft  so  that  the  safe  working  stress  in  tension  shall  be 
10,000  Ibs.  per  sq.  in.  and  in  shear  8000  Ibs.  per  sq.  in.  In 
this  case  M  =  240,000  and  7^  =  300,000.  Substituting  these 


114  STRENGTH  OF  MATERIALS 

values  in  Equation   (63)   gives  as  an  equivalent  bending 
moment, 


Me  =  ?^9+i  V2400002+3000002; 

=  120000+192000  =  312000  in.-lbs. 
The  diameter  of  shaft  is  found  from  Equation  (68)  thus: 


10.2X312000 
-J50TO—  =  6.82  m. 


The  equivalent  twisting  moment  is  found  from  Equation  (64) 
or 


Te  =  V  2400002+3000002  =  384000  in.-lbs. 
Substituting  this  value  in  Equation  (70)  gives 


0,^.1X300000  . 

d  =  ^' 8000 5'76lnS- 


This  shows  that  the  diameter  of  shaft  required  is  the  larger 
value,  or  6.82  ins. 

No  attempt  has  been  made  to  derive  Equations  (64)  and 
(63),  as  the  theory  involved  is  almost  too  complicated  for  a 
text  of  this  character.  For  line  shafts  carrying  gears  or 
pulleys  a  number  of  empirical  formulae  have  been  suggested. 
Table  XV  gives  values  recommended  by  the  A.  &  F.  Brown 
Co.,  of  New  York.  It  will  be  noted  that  the  commercial 
size  of  shafting  is  given  in  sixteenths,  and  advances  by  |-in. 
increments.  The  fiber  stress  used  in  this  table  •  is  about 
3500  to  4000  Ibs.  per  sq.  in. 


SHAFTING 


115 


|a 

S  s  •  s 

H    2& 

]     W    O 

^ 


o3       o3 

I  -§ 
5  -g 


1 


8  fl«2- 


s& 


COINOOOOOOOOOOOOOOO 
t-^<N^»OGOOCOT-iCqiO^COOOcOQOO 


CO  I-H  <M  O  <N 

^(XO(NO5 


CDOCO'O'OO^O'O'-OOiOOOiOO 
i—  ii-nfMCOiOCOOOCOOOOOifNt^iO 


OCOQOGOOOOfNcOOOO 

T}iiOCOO'>0<Mr-HT-HTTOQ;S; 

i—  t  i—  ((NCO'*»OCOOO 


iOi—  i  O  O  i—  i'—  lO^-i^fNiOO 
T-I  (N  CO  Tt»  iO  00  <N  t^.  CO  r-t  O  T-  i 


T-iiOOI^T^fNTti 
T-I  T-H  (N  (N  CO  Tt<  O 


ngi- 
and 


)  « 

r§ 

.2  fl 


I 


"SI 

oi  o> 
STJ 

8-9 
^1 

•ga 
|l 


5 

II 


jsl 


li 

o"S 

O    8J 

.2& 

a-° 

i1! 

If-! 

5S: 


116 


STRENGTH  OF  MATERIALS 


PROB.  122.  The  maximum  twisting  moment  in  a  crank  shaft 
is  150,000  in.-lbs.  and  the  bending  moment  is  120,000  in.-lbs. 
Find  the  proper  diameter  of  shaft,  using  a  safe  shearing  stress  of 
6000  Ibs  per  sq.  in.  and  a  bending  stress  of  8000  Ibs.  per  sq.  in. 

ART.  40.     PRACTICAL  PROBLEMS 

PROB.  123.  Fig.  74  shows  the  arrangement  of  the  crank  shaft 
for  a  vertical  side  crank  engine  driving  a  direct-connected  generator. 


FIG.  74. 

Diameter  of  cylinder  is  24  ins.  and  stroke  is  18  ins.  Maximum 
unbalanced  steam  pressure  is  100  Ibs.  per  square  in.  Find  the  reac- 
tions Ri  and  R2  for  up  and  down  strokes. 

PROB.  124.  Construct  the  bending  moment  diagrams  for  both 
up  and  down  strokes,  and  determine  the  maximum  bending  moment. 

PROB.  125.  Draw  the  vertical  shear  diagrams  for  both  up  and 
down  strokes. 


SHAFTING  117 

PROB.  126.  Determine  the  maximum  twisting  moment,  assum- 
ing that  the  maximum  tangential  thrust  on  the  crank  is  equal 
to  the  maximum  thrust  on  the  piston. 

PROB.  127.  Determine  the  equivalent  bending  and  twisting 
moments  in  inch-pounds. 

PROB.  128.  Using  S  as  10,000  Ibs.  per  sq.  in.  in  tension  and  as 
8000  in  shear,  determine  proper  diameter  of  shaft. 

PROB.  129.  Using  the  diameter  found  in  Prob.  128  figure  the 
probable  weight  of  the  shaft. 

PROB.  130.  With  the  assumed  diameter  figure  the  maximum 
fiber  stress,  taking  into  account  the  weight  of  the  shaft. 

PROB.  131.  Determine  the  diameter  of  the  crank  pin,  assuming 
that  the  length  of  the  pin  equals  1|  times  the  diameter. 


CHAPTER  VIII 
COLUMNS 

ART.  41.    STRESSES  IN  COLUMNS 

WHEN  a  short  specimen  is  placed  under  a  compressive 
load,  the  specimen  tends  to  fail  in  shear.  For  example,  a 
cast-iron  specimen  of  the  form  shown  in  Fig.  8  will  fail  by 
shearing  along  the  plane  A  B.  This  plane  makes  an  angle  of 
45°  plus  the  angle  of  repose  of  the  material,  with  the  vertical. 
As  the  length  of  the  specimen  relative  to  the  diameter  is 
increased,  this  angle  of  shear  becomes,  less  marked,  and 
eventually  the  specimen  will  fail  by  buckling  of  the  fibers. 
A  specimen  which  has  a  length  of  at  least  ten  times  the  least 
cross-sectional  dimension  of  the  specimen  is  referred  to  as  a 
column  or  strut. 

The  action  of  the  internal  stresses  in  a  column  is  somewhat 
similar  to  a  beam,  with  this  difference — that  in  addition  to 
the  bending  stress  there  is  also  present  a  direct  compressive 
stress.  As  stated  before,-  if  the  specimen  is  very  short,  the 

p 
unit  stress  under  a  load  of  P  pounds  equals  S  —  -r.    This  unit 

compressive  stress  continues  to  exist  as  the  length  is  increased, 
but  there  is  an  additional  stress  set  up  due  to  bending,  so 
that  the  safe  load  that  can  be  carried  by  a  column  gradually 
decreases  for  a  given  cross-section  as  the  length  increases. 
A  column  will  bend  in  a  plane  parallel  to  the  shorter  side 
of  the  column.  Hence  to  make  a  column  equally  strong  in 
either  direction  it  is  essential  that  the  cross-section  be  sym- 

118 


COLUMNS 


119 


metrical  with  respect  to  both  axes.    This  is  not  always  pos- 
sible, owing  to  the  difficulties  of  construction. 

Columns  are  usually  made  of  wood,  cast  iron,  steel,  or 
concrete.  The  load  carried  by  a  column  will  depend  upon 
how  the  ends  of  the  column  are  arranged.  Three  classes  of 
columns  are  recognized. 

I.  Columns  having  ends  fixed,  as  shown  in  Fig.  75. 

II.  Columns  having  one  end  fixed  and  the  other  end  free 
as  in  Fig.  76. 

III.  Columns  having  both  ends  free  or  pin  ended  as  shown 
in  Fig.  77. 


FIG.  75. 


FIG.  76. 


FIG.  77. 


Columns  in  Class  I  are  used  in  building  and  bridge  con- 
struction. The  column  in  such  cases  is  rigidly  secured  to 
beams  or  concrete  pedestals  or  similar  fixed  fastenings. 

Columns  in  Class  II  are  met  with  in  the  construction  of 
machines,  as,  for  example,  the  piston  rod  of  an  engine. 

Columns  in  Class  III  are  met  with  in  bridge  and  machine 
construction.  The  connecting  rod  of  a  steam  engine  is  a 
good  illustration  of  a  pin-ended  column. 

Experiment  has  shown  that  columns  of  Class  I  are  the 
strongest,  Class  II  is  second,  and  Class  III  the  weakest. 

If  the  column  is  loaded  eccentrically  the  safe  load  is 
materially  effected,  as  a  greater  bending  action  is  thereby 
induced. 

PROB.  132.  Calculate  the  theoretical  angle  of  shear  for  cast- 
iron  and  wood  specimens  when  subjected  to  a  direct  compressive 


120  STRENGTH  OF  MATERIALS 

stress.     The  angle  of  repose  of  these  materials  can  be  found  in  any 
reference  book  on  Mechanics. 

ART.  42.     RADIUS  OF  GYRATION 

The  moment  of  inertia  of  a  cross-section  is  given  by  the 
equation  I  =  Ar2,  where  A  is  the  area  of  the  cross-section  and 
r2  is  a  term  representing  the  average  of  the  squares  of  the 
distances  of  each  element  of  area  from  a  given  axis.  For 
example,  consider  the  case  of  an  I  beam;  it  is  apparent  that 
the  moment  of  inertia  can  be  figured  about  either  the  vertical 
axis — that  is,  an  axis  parallel  to  the  web  of  the  beam,  or 
about  a  horizontal  axis — that  is,  an  axis  parallel  to  the 
flanges  of  the  beam.  In  structural  work  it  is  usual  to  refer 
to  the  moment  of  inertia  about  the  vertical  axis  as  /2~2,  and 
about  the  horizontal  axis  as  /i-i.  These  two  values  are 
always  different,  and  in  the  case  of  a  column  the  lesser  value 
is  to  be  used.  When  this  lesser  value  is  used  the  quantity 
r  is  called  the  radius  of  gyration,  or 


(71) 


where  /  is  the  least  moment  of  inertia  and  A  the  area  of  the 
cross-section  in  inches.  For  example,  the  moment  of  inertia 
of  a  light  12-in.  I  beam  about  the  axis  1-1  is  215.8,  while 
about  the  axis  2-2  the  moment  of  inertia  is  9.5,  and  the  area 
of  the  section  is  9.26  sq.  ins.  Hence 

1.01. 
\A     \y.zt> 

For  a  circle 


d 

r= 


COLUMNS  121 

For  a  hollow  circle  of  outside  diameter  d  and  inside  diameter  d\ 


L,       •    •    (73) 


For  a  square  of  side  d 


For  a  hollow  square 


The  values  of  r  for  other  sections  can  readily  be  determined 
provided  the  moment  of  inertia  is  known.  No  attempt 
should  be  made  to  determine  the  value  of  r  independent  of 
the  moment  of  inertia. 

In  the  design  of  columns  the  least  moment  of  inertia  is 
to  be  used,  as  the  column  will  bend  in  a  plane  perpendicular 
to  the  axis  about  which  the  moment  of  inertia  is  the  least. 

PROB.  133.  Find  the  radius  of  gyration  for  a  hollow  circular 
section  whose  outside  diameter  is  12  ins.  and  inside  diameter  9  ins. 

PROS.  134.  Find  the  least  radius  of  gyration  for  a  light  12-in, 
channel  beam.  (Refer  to  Cambria  or  Carnegie  handbook.) 


FIG.  78. 

PROB.  135.    Fig.  78  shows  the  cross-section  of  a  plate  and  angle 
column.     The    angles    are    4X3XA-     The    plate   is    14  X  A    in. 


122 


STRENGTH  OF  MATERIALS 


Calculate  the  radius  of  gyration  about  the  axis  2-2. 
structural  handbook  for  moment  of  inertia  of  the  angles. 


Refer  to 


ART.  43.     RANKINE  COLUMN  FORMULA 

It  is  difficult  to  derive  a  rational  formula  for  columns 
such  as  was  done  in  the  case  of  beams.  The  result  has  been 
that  there  are  many  empirical  formulae  now  in  use  for  the 
design  of  columns.  The  formula  developed  by  Rankine  is 
one  commonly  used  by  engineers. 

Fig.  79  shows  a  column  under  the  action  of  a  vertical  load 
P,  which  tends  to  deflect  the  neutral  axis  of  the  column  from 
the  vertical  line  y-y  to  the  curved  line 
y-m-y.     The  vertical  load  P  produces  a 


of  -j  pounds 


uniform  compressive  stress 

on  the  cross-section  of  the  column.  Let 
this  stress  be  represented  by  AB,  Fig.  80, 
and  be  designated  Sc,  and  let  the  line  AC 
represent  the  least  dimension  of  the 
column.  It  is  evident  that  as  the 


BN 


FIG.  80. 


column  deflects,  the  fibers  on  the  side  EF  are  put  in 
compression,  due  to  the  bending  action,  and  the  fibers 
on  the  side  GH  are  put  in  tension.  Let  this  bending 
stress  be  designated  &>.  The  greater  the  deflection  of  the 


COLUMNS  123 

column  the  greater  will  be  the  stress  &.  Let  d  =  the  lateral 
deflection  of  the  column  =  mn.  The  bending  moment  on 
the  section  AB  due  to  the  vertical  load  P  equals  PXmn,  or 
PXd,  and  the  unit  stress  on  the  outer  fibers  of  the  column 

due  to  this  bending  moment  equals  $6  =  —  =  —  =-.     This 

bending  stress  varies  from  a  maximum  at  the  outermost 
fiber  of  the  column  to  zero  at  the  neutral  axis.  Let  the  line 
GH,  Fig.  80,  represent  the  variation  in  this  fiber  stress. 
Then  it  is  apparent  that  the  total  stress  on  the  outermost 
fiber  at  the  point  o  is  equal  to  the  direct  unit  compressive 
stress  due  to  bending.  Let  £  =  the  maximum  fiber  stress 
at  the  point  o,  then, 


but 


Q       P  A       Q       Pdc 

'  ^6== 


A'  7 

or 


(76) 


It  can  be  proven  that  the  deflection  d  is  proportional  to  the 
square  of  the  length  of  the  column,  or  dccL2,  hence  d  =  KL2 
where  K  is  a  constant.  From  article  42,  7  =  Ar2,  now  sub- 
stituting these  values  in  Equation  (76)  gives 


=  P    PKctf 
Let  Kc  =  5,  then, 


or 

(78) 


124 


STRENGTH  OF  MATERIALS 


where  S  equals  the  safe  unit  compressive  stress,  P  equals  the 
total  load  in  pounds,  A  equals  the  area  of  the  cross-section  in 
square  inches,  q  equals  a  constant  (see  Table  XVI),  L  equals 
the  length  of  the  column  in  inches,  and  r  equals  the  radius 
of  gyration.  If  the  dimensions  of  the  column  are  known 
the  safe  load  is  found  from  the  equation 

P  =  -^ro  (79) 


For  example,  consider  the  case  of  a  wooden  column 
8X8  ins.  in  cross-section  which  is  12  ft.  long  and  has  square 
ends.  Let  the  safe  unit  compressive  stress  equal  1200. 
The  area  A  in  this  case  equals  8X8  =  64  sq.  ins.  From 
Table  XVI  for  a  wooden  column  having  square  or  fixed  ends 
I/  =  12  ft.  =  144  ins.  For  a  square  section 


P  = 


AS 


therefore 


64X1200 


76800 


1       1442      2.3 


=  33,400  Ibs. 


3000  5.33 


If  a  block  of  wood  of  the  same  cross-section  were  under  a 
direct  compressive  stress  the  safe  load  would  be  64X1200  = 
76,800  Ibs.  in  place  of  33,400  Ibs.  as  above,  which  clearly 
shows  the  rapid  decrease  in  load  as  the  length  of  the  column 
increases. 

TABLE  XVI 

VALUES  OF  CONSTANT  q 


Material. 

Both  Ends 
Fixed. 

One  Fixed  and 
One  Pin  End. 

Both  Ends 
Round. 

Timber. 

1/3000 

2/3000 

4/3000 

Cast  iron. 

1/5000 

2/5000 

4/5000 

Wrought  iron  
Steel 

1/35000 
1/25000 

2/35000 
2/25000 

4/35000 
4/25000 

COLUMNS  125 

PROB.  136.  Find  the  safe  load  that  can  be  placed  on  a  hollow 
yellow-pine  column  12x12  outside  dimensions  and  8x8  inside 
dimensions;  length  of  column  16  ft.  Assume  that  the  ends  are 
fixed. 

PROB.  137.  Derive  an  expression  for  the  diameter  of  a  round 
.  column  in  terms  of  the  load  P,  the  area  A,  the  length  L,  and  the 
radius  of  gyration  r.  Use  Equation  (79)  as  a  basis. 

AKT.  44.     CAST-IRON  COLUMNS 

Cast-iron  columns  are  frequently  used  in  building  con- 
struction owing  to  their  cheapness,  strength  and  ease  of 
casting.  The  bases  are  cast  integral  with  the  column,  so 
that  they  can  readily  be  set  in  place.  The  cast-iron  column 
is  supported  on  a  stone  or  concrete  pier  into  which  have  been 
set  holding-down  bolts  to  prevent  the  column  from  moving 
in  a  lateral  direction. 

The  Cambria  Steel  Co  recommend  the  use  of  the  follow- 
ing formula  for  determining  the  safe  load  in  pounds  per 
square  inch  that  can  be  carried  by  a  cast-iron  column: 


(80) 


where  P  equals  the  safe  load  in  pounds  per  square  inch  of 
area,  L  equals  the  length  of  the  column  in  inches,  and  d  equals 
the  outside  diameter  in  inches.  For  example,  consider  the 
case  of  a  cast-iron  column  6  in.  in  diameter  and  12  ft.  long; 
here  L  -  144  and  L2  =  20,736  ;  d  =  6  and  d2  =  36. 
Hence 

„        10000        ro™iu 
P  =  -  20736  per  sq  m* 

+28800 

The  area  of  the  column  equals  .7854  X62  =  28.27  sq.  ins. 
Hence  the  total  load  =  5800X28.27  =  164,000  Ibs. 


126  STRENGTH  OF  MATERIALS 

For  the  same  data  Rankine's  formula  gives 

AS         28.27X10000 
P  --       —     =       -  —  =  10°  00°  (aPProx-)  • 


These  two  values  show  the  wide  range  in  values  given  by 
various  formulae.  For  a  solid  cast-iron  column  d  inches  in 
diameter  Equation  (79)  becomes 


.    .    .    (81) 


L2  L2X16* 


This  equation  can  be  reduced  to  a  biquadratic,  but  the 
author  recommends  that  the  known  n'umerical  values  be 
substituted  and  the  equation  reduced  to  its  simplest  form. 
It  will  be  found  that  the  resulting  equation  will  always  be  of 
the  form 

Q,       .....     (82) 


which  can  readily  be  solved  by  completing  the  square.  For 
a  hollow  cast-iron  column  the  problem  is  a  little  more  com- 
plex. For  example,  let  it  be  required  to  find  the  thickness 
of  a  hollow  cast-iron  column  20  ft.,  in  length  to  carry  a  load 
of  164,000  Ibs.,  the  outside  diameter  being  10  ins.  Assume 
the  ends  fixed  and  use  a  safe  fiber  stress  of  10,000  Ibs.  per 
square  inch.  Let  d  equal  the  unknown  inside  diameter; 
then  the  area  of  the  column  equals 

TT  /ir  100+d2 

A=j(100  —  d2)  and    r2  =  —  -^  —  . 

Substituting  these  values  in  Equation  (79)  gives 

,  j(100-d2)  X  10000     j(100-d2)  X  10000 


p_ 


L2  1      57600X16  184.3 


r2  5000    (100+d2) 


COLUMNS  127 

Now  reduce  the  right-hand  member  to  the  form 


384.  3+d2  284.  3+d2 

but 

P=  164000, 

hence 

164000     7854(1QQQQ-rf4) 
284.  3+d2      ' 

d4+20.9d2-4059  =  0; 


Hence  the  thickness  of  the  metal  =  --  x—  =  1.3  ins. 

& 

Rankine's  formula  is  gradually  being  replaced  by  the 
straight-line  formula,  so  called  because  when  values  are  sub- 
stituted in  the  formula  and  the  results  plotted  to  scale  the 
resulting  curve  is  a  straight  line.  For  square-end  timber 
columns  of  long-leaf  yellow  pine  the  New  York  building 
laws  specify  the  Equation, 


(83) 


where  D  is  the  least  transverse  dimension. 

For  cast-iron  columns  the  straight-line  formula  is  given  as, 

-^=10000-40^,       .....     (84) 

where  —  is  the  ratio  of  the  length  to  the  least  radius  of 
gyration. 


128  STRENGTH  OF   MATERIALS 

For  steel  columns  New  York  City  requires  the  following: 
^=15200-58- (85) 

A.  T 

PROB.  138.  Find  the  safe  load  by  New  York  law  on  a  6x8 
yellow-pine  post  12  ft.  long. 

PROB.  139i  Figure  the  safe  load  for  Prob.  138  by  using  Ran- 
kine's  formula. 

PROB.  140.  A  cast-iron  column  18  ft.  long  is  to  carry  200,000 
Ibs.  Assuming  the  ends  fixed,  find  the  outside  diameter  if  the 
inside  diameter  is  8  ins.,  so  that  the  maximum  fiber  stress  shall  not 
exceed  10,000  Ibs.  per  sq.  in. 

ART.  45.     STRUCTURAL  COLUMNS 

For  structural  work  columns  are  made  of  steel  I  beams  or 
combinations  of  the  other  rolled  shapes.  When  an  I  beam  is 
used  as  a  column  the  safe  load  is  determined  by  using  any 
of  the  standard  column  formulae,  but  in  every  case  the  least 
radius  of  gyration  must  be  used.  Fig.  81  shows  the  arrange- 
ment of  plates,  angles,  channels,  and  Z  bars  in  various  types 
of  columns.  Fig  81a  and  816  show  two  forms  of  plate  and 
angle  column.  To  determine  the  safe  load,  the  moment 
of  inertia  must  be  taken  about  the  1-1  axis.  In  figuring  the 
moment  of  inertia  it  is  usual  to  neglect  the  rivet  holes.  Fig. 
81c  shows  a  form  of  latticed  channel  column.  In  figuring 
the  safe  load  the  moment  of  inertia  of  the  channels  is  used, 
and  it  is  assumed  that  the  lattice  work  merely  ties  the  columns 
together  and  prevents  lateral  deflection.  Figs.  Sid  and  Sle 
show  forms  of  plate  and  channel  column.  Fig.  8 If  shows  a 
form  of  Z-bar  column. 

In  designing  a  structural  column  it  is  desirable  to  so 
space  the  various  elements  that  the  moments  of  inertia  about 
the  axes  1-1  and  2-2  will  be  as  near  equal  as  possible. 

Cambria  Steel  Co.  suggest  using  Gordon's  formula  for 
determining  the  safe  load  on  steel  columns,  using  a  factor  of 
safety  of  four,  so  that  the  unit  stress  in  Equation  (86),  Art. 
46,  will  equal  12,500. 


COLUMNS 


129 


For  example,  a  steel  column  of  the  form  shown  in  Fig.  Sid 
is  made  up  of  two  12-in.  Cambria  channels  weighing  25  Ibs. 
per  foot  and  two  steel  plates  14  Xj  in.  By  reference  to 
Cambria  handbook  it  is  found  that  the  moment  of  inertia  is 
the  least  about  the  axis  2-2  and  equals  524.3.  In  case  the 
value  of  the  moment  of  inertia  is  unknown  it  may  be  figured 
as  follows:  Let  Fig.  82  represent  the  cross-section  of  the 


I 


r-r^ 


{<- -11- ->! 


*  n 

i 

1 

1 

^ 

< 



r 

:._ 

__! 

i 

Y           12          Y| 
11"             ^. 

1 

d 

Y 


_J 


v 


FIG.  81. 


column  and  assume  the  flanges  to  be  of  uniform  thickness 
equal  to  the  average  thickness  of  the  top  and  bottom  of  the 
flanges.  This  average  thickness  equals  0.5  in.  Now  divide 
the  section  above  the  axis  2-2  into  the  rectangles  I,  II,  III, 
IV,  and  V,  having  dimensions  as  shown  in  Fig.  82.  The 
computations  for  the  moment  of  inertia  are  given  in  Table 
XVII.  The  moment  of  inertia  of  the  entire  figure  equals 


130 


STRENGTH  OF  MATERIALS 


twice  this  amount,  as  the  section  is  symmetrical  with  respect 
to  the  axis  2-2.  It  will  be  noted  that  the  value  of  I  as  com- 
puted equals  563.78,  while  that  given  in  Cambria  is  524.3. 
This  difference  is  due  to  the  fact  that  the  channels  are  not  set 
flush  with  the  edge  of  the  plate.  In  making  calculations  for 
I  the  exact  dimensions  of  the  column  are  to  be  used. 


FIG.  82. 
TABLE  XVII 


Mark. 

b 

h 

hi 

h* 

/«» 

fc  -  hi* 

b(h*-hi*) 

I 

.5 

7 

0 

343 

0 

343 

171.50 

II 

.5 

7 

3.95 

343 

61.63 

281.37 

140.68 

III 

11 

4.34 

3.95 

81.75 

61.63 

20.12 

221.32 

IV 

.5 

7 

3.95 

343 

61.63 

281.37 

140.68 

V 

.5 

7 

0 

343 

0 

343 

171.50 

3 

845.68 

Above 

axis 

281.89 

Below 

axis 

281.89 

la—  2  — 

563.78 

COLUMNS  131 

After  the  least  value  of  7  has  been  determined  the  radius 

of  gyration  is  found  from  the  formula  r  =  \j-r>     In  this  case 

1563.78 
r=\    287  Based  on  the  value  given  in  Cambria 

r=4.27. 

Using  Gordon's  formula  and  assuming  length  of  column 
equals  20  ft.,  the  total  load  that  can  be  carried  by  the  column 
equals 

„     ^X12500     or    P  =  _^7X12500_=330)0001bg> 


1+ 


(12L)2 
36000r2 


1+ 


36000X18.23 


L^y 


4L'__ 
0.677- 


0.513-> 


ft 


FIG.  83. 

Another  method  of  figuring  the  moment  of  inertia  of  a 
compound  section  is  as  follows:  Let  Fig/ 83  represent  the 
cross-section  of  a  plate  and  channel  column.  From  the 
handbook  the  moment  of  inertia  of  the  two  channels  about 
the  axis  1-1  equals  2X161.7-323.4.  The  channels  weigh 
30  Ibs.  per  ft.,  and  the  plates  are  14 Xf  in.  The  moment  of 
inertia  of  the  plate  relative  to  its  own  gravity  axis  equals 


132  STRENGTH  OF  MATERIALS 

14  X  - 3 

4  =.98.     This    moment    must    be    transferred  to  the 

axis  1-1  by  means  of  the  reduction  formula  explained  in 
Art.  23,  or 

7i-i  (of  plate)  =Ig+Ad2  =  . 98+ 10.5 X.3752  =427.7. 

Hence  the  total  moment  of  inertia  of  the  section  relative  to 
the  axis  1-1  equals. 

/i-i  for  two  12-in.  channels  =  2X16 1.7  =323.4 

1 1-1  for  two  14-in.Xt-in.  plates  =  2X427. 7    =  ^4; 

117o . o 

The  ratius  of  gyration  equals  \J-T- 

PROB.  141.  Find  the  safe  load  that  can  be  carried  by  a  light 
12-in.  I  beam  when  used  as  a  column.  Length  of  column  equals 
25  ft. 

PROB.  142.  A  column  of  the  form  shown  in  Fig.  81a  is  made  up 
of  four  angles  6x4x^  in.  and  of  a  12  xi  in.  web  plate.  If  the 
length  of  column  is  22  ft.,  find  the  safe  load.  *. 

PROB.  143.  Select  a  suitable  I  beam  to  be  used  as  a  column 
24  ft.  long  to  support  a  load  of  75,000  Ibs.  Assume  ends  fixed. 

ART,  46.  COLUMN  FORMULA 

In  designing  columns  the  engineer  must  so  proportion 
the  metal  that  the  column  will  support  a  maximum  load  with 
a  mimimum  of  metal.  This  means  selecting  the  elements 
of  the  column  so  that  the  moment  of  inertia  of  the  section 
shall  be  a  maximum  for  a  given  weight.  Many  formula 
have  been  suggested  to  simplify  the  calculations  for  columns. 
The  Cambria  Steel  Co.  recommend  Gordon's  formula  for 
steel  columns  with  fixed  ends,  which  is, 

50000 

r°        — 7ior\2»         (86) 


COLUMNS  133 

where  P  equals  the  maximum  load  in  pounds  per  square  inch 
which  the  column  can  support.  This  value  is  to  be  divided 
by  the  desired  factor  of  safety,  which  is  taken  as  four; 
L  equals  the  length  of  the  column  in  feet,  and  r  equals  the 
least  radius  of  gyration  of  the  section.  The  total  safe  load 
which  the  column  can  support  equals  the  load  P  times  the 
cross-sectional  area  of  the  column,  divided  by  the  desired 
factor  of  safety. 

Many  of  the  formulae  take  the  form  of  the  equation  of  a 

straight  line.     These  are  in  the  form  P—lA  —  C—j,  where 

A  and  C  are  constants,  L  equals  the  length  in  inches  and  r 
equals  the  least  radius  of  gyration.  For  steel  columns  with 
fixed  ends  the  building  laws  of  New  York,  Philadelphia, 
and  Boston  require  the  use  of  the  following  formulae: 

New  York 

P=  15200  -58^  ......  .    (87) 

Philadelphia 

P'-~  ......     (88) 


11000r2 
Boston 


fr 

(89) 


1+ 


20000r2 
The  American  Railway  Engineers  Association  recommend 

P=  16000  -70-       .....     (90) 

In  all  cases  P  equals  the  safe  unit  stress  in  pounds  per  square 
inch  of  cross-sectional  area  of  the  column.     It  is  customary 

to  limit  the  maximum  ratio  of  —  to  100  to  120. 

r 


134  STRENGTH  OF  MATERIALS 

Another  formula  which  is  sometimes  used  is  that  sug- 
gested by  Euler,  which  is 


(91) 


where  P  equals  the  load  in  pounds,  E  equals  the  modulus  of 
elasticity  of  the  material,  /  equals  the  least  moment  of  inertia 
of  the  section,  and  L  equals  the  length. 

PROS.  144.  Compare  the  safe  load  that  can  be  carried  by  a 
12-in.  steel  I  beam  when  used  as  a  column  20  ft.  long,  using  Equa- 
tions (87),  (88),  and  (89). 

PROB.  145.  Find  the  safe  load  that  can  be  carried  by  a  column 
of  the  form  and  dimensions  shown  in  Fig.  8ld.  Length  of  column  is 
24  ft.  Channels  weigh  30  Ibs.  per  lin  ft.  Plates  are  1  in.  thick. 
Figure  the  load  by  using  both  Equations  (79)  and  (90).  Compare 
the  results. 


ART.  47.     REVIEW  PROBLEMS 

PROB.  146.  By  use  of  Equation  (80)  find  the  safe  load  that 
can  be  placed  on  a  hollow  cast-iron  column  22  ft.  in  length.  Out- 
side diameter  of  column  is  14  ins.  and  the  inside  diameter  11  ins. 

PROB.  147.  A  hollow  cast-iron  column  supports  a  load  of  330,- 
000  Ibs.  The  column  is  20  ft.  in  length.  Find  the  thickness  of  the 
metal  if  the  outside  diameter  is  12  ins.  Use  Rankine's  formula. 

PROB.  148.  A  steel  column  is  made  up  of  two  15-in.  columns 
weighing  40  Ibs.  per  ft.,  and  two  steel  plates  20  ins.  wide  and  \  in. 
thick.  The  channels  are  placed  back  to  back.  Find  the  safe  load 
if  the  column  is  24  ft.  long.  Use  Equation  (87). 

PROB.  149.  The  diameter  of  the  cylinder  of  an  engine  is  24  ins. 
The  maximum  unbalanced  steam  pressure  is  120  Ibs.  per  sq.  in. 
Find  the  diameter  of  a  steel  piston  rod,  assuming  the  length  of  the 
rod  to  be  6  ft. 

PROB.  150.  Determine  the  diameter  of  the  connecting  rod  of 
the  engine  in  Prob.  149,  assuming  the  stroke  of  the  engine  to  be 


COLUMNS  .    135 

42  ins.  and  the  length  of  the  connecting  rod  to  be  9  ft.  (NOTE. — 
determine  the  maximum  thrust  on  the  connecting  rod.) 

PROB.  151.  A  9x12  in.  piece  of  yellow  pine  16  ft.  long  is 
used  as  a  column  with  fixed  ends.  Use  Equation  (83). 

PROB.  152.  Find  the  safe  load  that  can  be  carried  by  a  column 
of  the  cross-section  shown  in  Fig.  83.  Length  of  column  is  36  ft. 

PROB.  153.  Find  the  radius  of  gyration  of  a  column  whose 
section  is  made  up  of  four  5x3^x£  angles  and  one  plate  10  X? 
arranged  as  shown  in  Fig.  81a. 

PROB.  154.  Find  the  moment  of  inertia  of  the  column  section 
shown  in  Fig.  816.  The  angles  are  6  X4  x£.  The  flange  plates  are 
14  xf,  and  the  web  plate  is  14  Xf  in. 

PROB.  155.  In  Prob.  154  determine  the  least  radius  of  gyration 
of  the  section. 


CHAPTER   IX 


RIVETED  JOINTS 

ART.  48.     CAST-IRON  PIPE 

WHEN  a  cylindrical  vessel  is  subjected  to  an  internal 
pressure  of  P  pounds  per  square  inch,  the  required  thickness 
of  metal  is  found  from  the  formula, 


or 


=  2St, 

_PD 

2S° 


(92) 
(93) 


where  P  equals  the  internal  pressure  in  pounds  per  square 
inch,  D  equals  the  inside  diameter  in  inches,  t  equals  the  thick- 
ness of  the  metal  in  inches  and  S  equals  the  safe  working 
stress  in  pounds  per  square  inch. 


^ 

c 

3 

; 

f 

FIG.  84. 

In  a  cylindrical  or  spherical  vessel  the  internal  pressure 
acts  in  a  radial  direction,  or  normal  to  the  surface  of  the 
vessel.  In  Fig.  84  consider  one-half  of  the  pipe  filled  with  a 
solid  substance.  It  is  evident  the  pipe  will  tend  to  break 

136 


RIVETED  JOINTS  137 

along  the  lines  MN  and  EF.  The  total  pressure  tending  to 
burst  the  pipe  equals  the  unit  pressure  P,  times  the  projected 
area  LD  of  the  pipe,  or  the  total  pressure  equals  PLD.  The 
strength  of  the  metal  resisting  this  pressure  equals  2tLS, 
assuming  that  the  pipe  is  equally  strong  on  either  side.  For 
equilibrium  to  exist,  the  internal  pressure  and  the  strength 
of  the  metal  must  be  equal,  or, 


hence, 

=  2St. 


The  above  formula  is  applicable  only  when  the  inside 
diameter  of  the  pipe  is  at  least  ten  times  the  thickness  of  the 
metal. 

Large  water  and  soil  pipes  are  made  of  cast  iron.  Owing 
to  the  imperfections  due  to  casting  and  to  the  structure  of 
the  cast  iron  a  large  factor  of  safety  is  to  be  used  in  deter- 
mining the  thickness  of  metal.  This  is  usually  taken  from 
12  to  16. 

For  example,  consider  a  pipe  24  ins.  in  diameter  subjected 
to  an  internal  pressure  of  100  Ibs.  per  sq.  in.  The  tensile 
strength  of  cast  iron  is  20,000,  so  that  using  a  factor  of 
safety  of  12  gives  a  unit  working  stress  of  approximately 
1600  Ibs.  per  sq.  in.  Substituting  this  value  in  Equation 
(93)  gives 


PD     100X24=3  . 
~~2S  ~2Xl600~4l] 


The  transverse  section  of  a  pipe  is  twice  as  strong  as  the 
longitudinal  section.     The  force  tending  to  burst  the  pipe 

on  a  transverse  section  equals  — r- XP.     The  strength  of  the 


138 


STRENGTH  OF   MATERIALS 


metal  resisting  this  force  equals  the  area  of  the  cross-section 
of  the  metal  in  the  pipe,  or  -n-DtS,  hence 


or 


=  4St 


(94) 


Equation  (94)  can  be  used  to  determine  the  thickness  of  a 
sphere. 

PROB.  156.  A  cast-iron  water  pipe  36  ins.  in  diameter  is  sub- 
jected to  a  head  of  250  ft.  Find  the  thickness  of  the  metal,  using  a 
factor  of  safety  of  12. 

PROB.  157.  Find  the  safe  working  pressure  that  can  be  placed 
on  a  48-in.  cast-iron  water  pipe  if  the  thickness  of  the  metal  is  2  ins. 


ART.  49.     LAP  JOINTS 

The  simplest  method  of  joining  two  plates  is  by  means 
of  a  riveted  joint.  The  rivets  are  made  of  iron  or  steel,  and 
consist  of  an  upset  end  called  the  head,  and  a  long  cylindrical 
part  called  the  shank.  Rivets  in  general  are  placed  at  right 
angles  to  the  forces  tending  to  cause  them  to  fail. 


Fig.  85  shows  various  forms  of  rivet  points.  Fig.  85a  is 
called  the  cone  or  steeple  point;  Fig.  856  is  known  as  the 
button  head,  while  Fig.  85c  shows  a  form  of  countersink  rivet. 


RIVETED  JOINTS 


139 


In  punching  the  plate  the  rivet  hole  is  made  ^  in.  larger 
than  the  diameter  of  the  rivet. 

In  designing  a  riveted  joint  for  structural  work  only  the 
question  of  strength  need  be  considered,  but  in  the  case  of 
boilers,  tanks,  etc.,  the  joint  must  be  both  strong  and  water- 
tight. 

The  simplest  form  of  riveted  joint  is  that  shown  in  Fig. 
86,  where  one  plate  overlaps  the  other,  and  the  plates  are 
secured  in  place  by  a  single  row  of  rivets.  Such  a  joint  is 
referred  to  as  a  single-riveted  lap  joint.  This  joint  may  fail 


© 


© 


© 
© 


©      ©      <§ 


FIG.  86. 

in  one  of  several  ways.  First,  the  rivets  may  shear;  second, 
the  plate  may  tear  between  the  rivet  holes;  third,  the  rivets 
may  shear  the  plate;  fourth,  the  rivets  may  crush  the  plate  in 
front  of  the  rivet. 

The  pitch  of  a  riveted  joint  is  the  distance  from  center  to 
center  of  rivets  in  the  outer  course  (provided  there  are  several 
rows  of  rivets). 

The  distance  from  the  edge  of  the  rivet  hole  to  the  edge  of 
the  plate  is  called  the  margin. 

The  distance  between  the  edges  of  two  plates  which  over- 
lap is  termed  the  overlap. 

In  no  case  can  the  strength  of  a  riveted  joint  be  greater 


140  STRENGTH  OF  MATERIALS 

than  the  strength  of  the  solid  plate.  It  is  usual  to  investigate 
a  riveted  joint  for  failure  of  the  rivets  in  shear  and  for  failure 
of  the  punched  plate  in  tension.  The  efficiency  of  a  riveted 
joint  is  based  on  the  strength  of  the  solid  plate.  The  effi- 
ciency of  the  plote  is  the  ratio  of  the  strength  of  the  punched 
plate  to  the  strength  of  the  solid  plate. 

The  efficiency  of  the  rivets  is  the  ratio  of  the  strength  of 
the  rivets  to  the  strength  of  the  solid  plate.  In  design 
both  of  these  efficiencies  are  calculated  and  the  lesser  value 
used. 

For  example,  in  the  case  of  a  single-riveted  lap  joint,  as 
shown  in  Fig.  86,  let  P  equal  the  pitch  in  inches,  d  equal  the 
diameter  of  the  rivet  after  being  driven  or  the  diameter  of 
the  rivet  hole,  T  equal  the  tensile  strength  of  the  plate  in 
pounds  per  square  inch,  S  equal  the  shearing  strength  of  the 
rivets  in  pounds  per  square  inch,  and  t  equal  the  thickness  of 
the  plate  in  inches.  Then  the  strength  of  the  solid  plate,  for  a 
distance  equal  to  the  pitch,  is  TxPXt,  and  the  strength  of 
the  punched  plate  is  TX  (P— d)  Xt,  hence  the  efficiency  of  the 
plate  equals 

TX(P-d)     P-d 


TxPXt   '      P 
and  the  efficiency  of  the  rivets  equals 

ird2XS 


.'•(95) 


(96) 


TXPXt 


The  shearing  strength  of  the  rivets  is  about  three-quarters 
the  tensile  strength  of  the  plate,  or  S  =  %T.  Substituting 
this  value  in  Equation  (96)  gives  the  efficiency  of  the  rivets 
equal 


.59d2 


TXPXt     PXt' 


RIVETED  JOINTS  141 

If  the  plate  is  to  be  equally  strong  with  the  rivets,  the 
efficiency  of  plate  and  rivets  must  be  equal,  or, 


PXt~    P    ' 
or 

(P-d)=~  .......     (97) 

From  Equation  (97)  the  required  pitch  may  be  found  for  a 
given  thickness  of  plate  and  diameter  of  rivet. 

For  boiler  work  the  American  Society  of  Mechanical 
Engineers  recommend  the  following  method  of  determining 
the  efficiency  of  a  single-riveted  lap  joint;  let  A  equal  the 
strength  of  the  solid  plate  equal  PXTXt;  let  B  equal  the 
strength  of  the  plate  between  rivet  holes  equal  (P  —  d)XTXt; 
let  C  equal  the  shearing  strength  of  one  rivet  in  single  shear 
equal  NXSXa,  where  N  equals  the  number  of  rivets  in 
single  shear  in  a  unit  of  length,  S  =  the  shearing  strength  of  a 
rivet  in  single  shear  and  a  equals  the  area  of  the  rivet  after 
driving;  and  let  D  equal  the  crushing  strength  of  the  plate  in 
front  of  the  rivet  equals  dXtXc,  where  c  equals  the  crushing 
strength  of  the  plate.  To  find  the  efficiency  of  the  joint, 
calculate  the  values  of  B,  C,  and  D;  divide  the  lesser  value 
by  the  value  of  A,  and  the  quotient  will  be  the  required 
efficiency. 

For  example,  let  the  pitch  P=  If  in.,  t  =  J  ins.,  diameter  of 
rivet  hole  equal  H,  7"  =  55,000,  £  =  44,000,  and  C  =  95,000. 

Then  A  =PxTXt=  If  X55,OOOXi  =  22,343. 

B=(P-d)X77XMl.625-.6874)X55,OOOXi  =  12,890; 
C  =  ArX*SXa=lX44,OOOX.3712=  16,332; 
D  =  dXtXc  =  .6875  X  i  X  95,000  -  16,328. 

The  value  of  B  is  less  than  that  of  C  or  D,  hence  the  efficiency 

,         ...     B     12890     __ 
of  the  joint  =  T  =  o9^n  Pe 


142 


STRENGTH  OF  MATERIALS 


For  complete  analysis  of  this  method  the  student  is 
referred  to  Vol.  XXXVI  of  the  Transactions  of  the  A.S.M.E. 

In  the  case  of  a  double-riveted  lap  joint,  as  shown  in 
Fig.  87,  the  same  method  can  be  applied.  In  figuring  the 
value  of  C  in  this  case  N  is  equal  to  two,  as  there  will  be  two 


FIG.  87. 

rivets  in  single  shear,  and  likewise  D  =  2XdXtXc,  as  there 
are  two  rivets  tending  to  crush  the  plate. 

PROB.  158.  Find  the  efficiency  of  a  double-riveted  lap*  joint 
where  the  pitch  of  the  rivets  is  3  ins.,  the  diameter  of  the  rivets 
H  in.  Assume  S=  44,000,  C  =  95,000,  and  T  =  55,000. 


ART.  50.     BUTT  JOINTS 

Lap  joints  are  used  in  making  the  girth  seams  of  boilers 
and  tanks.  The  longitudinal  seams  are  formed  by  butting 
the  two  ends  of  the  plate  together  and  securing  them  in  place 
by  use  of  an  inside  and  outside  cover  plate,  as  shown  in  Fig. 
88.  This  is  referred  to  as  a  butt  joint.  In  Fig.  88  the  joint 
is  double  riveted,  in  Fig.  89  the  joint  is  triple  riveted,  and  in 
Fig.  90  the  joint  is  quadruple  riveted.  The  more  rows 
of  rivets  the  more  nearly  will  the  efficiency  of  the  joint  equal 
that  of  the  solid  plate.  Theoretically  a  rivet  should  be  twice 


RIVETED  JOINTS 


143 


as  strong  in  double  shear  as  in  single  shear,  but  practically 
such  is  not  the  case.  The  A.S.M.E.  recommend  the  following 
values  for  iron  and  steel  rivets  in  single  and  double  shear : 

Iron  rivets  in  single  shear  =38,000; 
Iron  rivets  in  double  shear  =76,000; 
Steel  rivets  in  single  shear  =44,000; 
Steel  rivets  in  double  shear  =88,000. 

In  calculating  the  efficiency  of  a  butt  joint  the  pitch  is 
taken  as  the  distance  between  centers  of  the  rivets  in  the 


FIG.  88. 


outer  course.  It  will  be  noted  that  in  the  double-riveted 
butt  joint  the  pitch  of  the  inner  row  is  one-half  that  of  the 
outer  row.  In  the  triple-riveted  butt  joint  the  pitch  of  the 
second  and  third  rows  is  one-half  that  of  the  outer  row,  or 
course. 

The  triple-riveted  butt  joint  is  commonly  used  in  making 
the  longitudinal  seam  of  the  steam  drum  of  water-tube 
boilers.  The  method  of  determining  the  efficiency  is  as 
follows ; 


144 


STRENGTH  OF  MATERIALS 


Let  A  equal  the  strength  of  the  solid  plate  equal  PXtX  T; 
B  equal  the  strength  of  the  plate  between  rivet  holes  in 
the  outer  course  equal  (P  —  d)XtXT; 


FIG.  89. 


FIG.  90. 


C  equal  the  shearing  strength  of  four  rivets  in  double 
shear,  plus  the  shearing  strength  of  one  rivet  in  single 
shear  equal  NXSXa+nXsXa; 


RIVETED  JOINTS  145 

D  equal  the  strength  of  the  plate  between  rivet  holes 
in  the  second  row,  plus  the  shearing  strength  of  one 
rivet  in  single  shear  in  the  outer  row  equal  (P  —  2d) 
XtXT+nXsXa; 

E  equal  the  strength  of  the  plate  between  rivet  holes  in 
the  second  row,  plus  the  crushing  strength  of  the 
butt  strap  in  front  of  one  rivet  in  the  outer  row  equal 
(P-2d)XtXT+dXbXc,  where 
S  equals  the  strength  of  rivet  in  double  shear; 
N  equals  the  number  of  rivets  in  double  shear  in  the  unit 
of  length  and  6  equal  the  thickness  of  the  butt  strap. 
The  efficiency  of  the  joint  is  determined  by  finding  the 
values  of  B,  C,  D,  and  E  and  dividing  the  least  of  these  values 
by  A.     For  example,  in  a  triple-riveted  butt  joint  with  inside 
and  outside  cover  plates  or  butt  straps,  the  thickness  of  plate 
is  §  in.,  P  =  6.5  ins.,  6  =  1%  ins.,  d  =  ft  ins. 

Here  A  =  PX  IX  T  =  6.5  X.  375X55,000  =134,062; 

B=  (P-d)  XtXT=  (6.5-.  8125)  X.  375X55,000  = 
117,304; 


44,000  X.  5185  =  205,326; 

D=(P-2d)X*X!T+nXsXa=(6.5-2X.8125)X.375 
X55,000+1X44,OOOX.  5185  =  123,360; 

E=(P-2d)XtX  T+dXbXc 
=  (6.5-2X.8125)X.375X55,000+.8125X.3125X 
95,000=124,667. 

Here  the  value  of  B  is  the  least,  hence  the  efficiency  equals 
B     117304 


In  structural  work  butt  joints  are  used  to  join  together 
various  sections  of  beams,  girders,  columns,  etc.  In  such 
cases  a  sufficient  number  of  rivets  must  be  provided  to  take 
care  of  the  vertical  shear  at  any  given  point.  The  Cambria 


146  STRENGTH  OF  MATERIALS 

Steel  Co.  suggest  the  following  rules  for  rivet  spacing  in 
bridge  and  structural  work: 

"  The  pitch  or  distance  from  center  to  center  of  rivets 
should  not  be  less  than  three  diameters  of  the  rivet.  In 
bridge  work  the  pitch  should  not  exceed  6  ins.,  or  sixteen 
times  the  thickness  of  the  thinnest  outside  plate,  except  in 
special  cases  hereafter  noted.  In  the  flanges  of  beams  and 
girders  where  plates  more  than  12  ins.  wide  are  used,  an 
extra  line  of  rivets  with  a  pitch  not  greater  than  9  ins.  should 
be  driven  along  each  edge  to  draw  the  plates  together. 

"  At  the  ends  of  compression  members  the  pitch  should 
not  exceed  four  diameters  of  the  rivet  for  a  length  equal  to 
twice  the  width  or  diameter  of  the  member. 

"  In  the  flanges  of  girders  and  chords  carrying  floors,  the 
pitch  should  not  exceed  4  ins. 

11  For  plates  in  compression  the  pitch  in  the  direction  of 
the  line  of  stress  should  not  exceed  sixteen  times  the  thickness 
of  the  plate,  and  the  pitch  in  a  direction  at  right  angles  to  the 
line  of  stress  should  not  exceed  thirty-two  times  the  thickness, 
except  for  cover  plates  of  top  chords  and  end  posts,  in  which 
the  pitch  should  not  exceed  forty  times  their  thickness. 

"The  distance  between  the  edge  of  any  piece  and  the 
center  of  the  rivet  hole  should  not  be  less  than  \\  ins.  for 
f  in.  and  |-in.  rivets,  except  in  bars  less  than  2J  ins.  wide; 
when  practicable  it  should,  for  all  sizes,  be  at  least  two 
diameters  of  the  rivet  and  should  not  exceed  eight  times  the 
thickness  of  the  plate. 

"  Minimum  spacing  is  generally  used  in  pin  plates,  at 
ends  of  columns,  girders,  etc. 

"  In  figuring  clearance  of  rivets  for  special  cases,  allow 
|  in.  in  addition  to  diameter  of  head.'7 

PROB.  159.  In  a  quadruple-riveted  joint  P  =  15  ins.,  t  =  %  in., 
b  =  Tg  in.,  d  =  H  in.  Find  the  values  of  A,  B,  C,  and/).  Determine 
the  efficiency  of  the  joint. 


RIVETED  JOINTS  147 


ART.  51.    APPLICATION  TO  BOILERS 

The  formula  for  determining  the  thickness  of  a  pipe  is  not 
applicable  to  a  vessel  which  is  made  up  of  rolled  plates  joined 
together  by  riveted  joints.  In  such  cases  the  efficiency  of 
the  joint  bears  an  important  part  in  fixing  the  thickness  of 
metal.  Various  formula?  have  been  suggested  from  time  to 
time,  and  the  one  now  recommended  by  the  A.S.M.E.  is 
the  following  : 


flXF.S.   ' 

where  T.S.  equals  ultimate  tensile  strength  stamped  on  shell 

plates,  pounds  per  square  inch  ; 
t  equals  minimum  thickness  of  shell  plates  in  the 

weakest  course  in  inches; 

E  equals  efficiency  of  the  longitudinal  joint  or  of 
ligaments  between  two  holes  (whichever  is  the 
least)  ; 
R  equals  the  inside  radius  of  the  weakest  course 

of  the  shell  or  drum,  in  inches; 
F.S.  equals  the  factor  of  safety,  or  the  ratio  of  the 
ultimate  strength  of  the  material  to  the  allow- 
able stress; 
P  equals  the  allowable  working  steam  pressure 

in  Ibs.  per  sq.  in. 

For  new  boilers  the  factor  of  safety  is  to  be  taken  as  five. 
This  formula  is  derived  by  considering  that  efficiency 

eauals  -  The  output  of  the  riveted  joint  is  holding 

input 

the  pressure  P,  against  a  given  radius  R,  or  equals  2PR. 
The  input  is  the  strength  of  the  punched  plate  equals  2tSt 
for  a  unit  of  length,  hence 

2PR    PR 


E  = 


2tSt      tSt ' 


148  STRENGTH  OF  MATERIALS 

but 


hence 

PXflXF.S. 

*XT.S. 
or 

T.S.XIXff 


The  required  thickness  of  plate  is  given  by  the  formula 

PXflXF.S. 
T.S.XE 

For  example,  consider  a  42-in.  steam  drum  such  as  is 
used  on  a  B.  &  W.  water-tube  boiler.  Assume  boiler  pressure 
equal  125  Ibs.  per  sq.  in.  gauge,  and  assume  the  longitudinal 
seam  to  be  a  triple-riveted  butt  joint  having  an  efficiency  of 
85  per  cent.  The  tensile  strength  of  the  plate  is  55,000  Ibs. 
per  sq.  in.  As  stated  above  in  new  installations  a  factor  of 
safety  of  five  is  to  be  used,  hence  the  thickness  in  this  case 
equals 

125X21X5  . 

=-28 


From  Equation  (99)  it  is  noted  that  the  lower  the  efficiency 
of  the  joint  the  greater  will  be  thickness  of  metal  required. 
In  large  fire-tube  boilers  of  the  locomotive  type  it  is  desirable 
to  use  a  joint  having  as  high  an  efficiency  as  possible. 

EXAMPLE.  A  locomotive  boiler  has  a  maximum  inside 
diameter  of  68  ins.  The  maximum  steam  pressure  is  220  Ibs. 
per  sq.  in.  The  longitudinal  seam  is  a  triple-riveted  butt 
joint  with  an  efficiency  of  87  per  cent.  Tensile  strength 
of  the  shell  plate  is  56,000  Ibs.  per  sq.  in.  The  thickness  of 


RIVETED  JOINTS  149 

the  shell  is  xJ  in.     Determine  the  factor  of  safety.     Using 
Equation  (99)  and  transposing  the  terms  gives, 


4.4. 


PXR 

55000X11X.87 


220X16X34 
In  locomotive  work  a  factor  of  safety  of  4.5  is  permissible. 

PROB.  160.  A  Scotch  marine  boiler  is  16  ft.  in  diameter,  and 
carries  a  pressure  of  175  Ibs.  per  sq.  in.  Using  a  shell  plate  having  a 
tensile  strength  of  56,000  Ibs.  per  sq.  in.,  find  the  thickness  of  the 
shell.  Longitudinal  seam  is  a  triple-riveted  butt  joint  with  an 
efficiency  of  88  per  cent. 

PROB.  161.  An  air  tank  carries  a  pressure  of  80  Ibs.  per  sq.  in. 
The  tank  is  48  ins.  in  diameter.  Longitudinal  seam  is  a  double- 
riveted  butt  joint  having  an  efficiency  of  72  per  cent.  Find  the 
proper  thickness  of  shell  plate  if  tensile  strength  of  plate  is  50,000 
Ibs.  per  sq.  in. 

ART.  52.     PRACTICAL  PROBLEMS 

PROB.  162.  Find  the  thickness  of  a  16-in.  cast-iron  standpipe 
which  is  subjected  to  a  head  of  water  of  250  ft.  Assume  a  steady 
stress. 

PROB.  163.  Find  the  pressure  required  to  burst  a  cast-iron 
cylinder  12  ins.  in  diameter  and  1.5  ins.  thick. 

PROB.  164.  Given  a  double-riveted  lap  joint.  Thickness  of 
plate  equals  \  in.  Diameter  of  rivets  equals  f  in.  The  strength 
of  the  punched  plate  to  equal  the  strength  of  the  rivets.  Assume 
Ss  =  f&.  Find  the  pitch  of  the  rivets  and  the  efficiency  of  the  joint. 

PROB.  165.  Diameter  of  a  steel  boiler  is  65  ins.  The  thickness 
of  the  shell  is  H  in.  Steam  pressure  is  180  Ibs.  per  sq.  in.  gauge. 
If  the  efficiency  of  the  longitudinal  seam  is  87  per  cent,  find  the  fac- 
tor of  safety. 

PROB.  166.  In  a  single-riveted  lap  joint  the  pitch  is  2|  in., 
diameter  of  rivets  is  f  in.  Diameter  of  shell  is  36  in.  If  the  steam 


150  STRENGTH  OF  MATERIALS 

pressure  is  100  Ibs.  per  sq.  in.,  find  the  unit  shearing  stress  on  the 
rivets,  and  the  unit  tensile  stress  on  the  plate.  The  lap  joint 
forms  the  girth  seam. 

PROB.  167.  A  Scotch  marine  boiler  is  16  ft.  in  diameter.  Longi- 
tudinal seam  is  a  triple-riveted  butt  joint  having  an  efficiency  of 
85  per  cent.  Maximum  steam  pressure  is  125  Ibs.  per  sq.  in. 
Find  thickness  of  the  shell. 

PROB.  168.  In  a  quadruple-riveted  joint,  the  pitch  of  the  rivets 
in  the  outer  row  is  18  ins.,  thickness  of  plate  is  H  in.,  thickness  of 
butt  straps  is  ^  in.,  diameter  of  rivets  is  1|  in.,  pitch  in  the  second 
row  is  9  ins.,  and  in  the  third  and  inner  rows  is  4|  ins.  Tensile 
strength  of  the  plate  is  55,000  Ibs.  per  sq.  in.  Draw  the  riveted 
joint  to  scale  and  figure  the  efficiency  of  the  joint.  Width  of  inner 
butt  strap  is  26  ins.  and  of  outer  butt  strap  is  12|  ins. 

PROB.  169.  In  a  triple-riveted  butt  joint  the  pitch  in  the  outer 
row  is  8  ins.  Thickness  of  plate  is  |  in.  and  of  butt  straps  is  re  in. 
Diameter  of  rivets  is  f  in.  Diameter  of  boiler  is  42  ins.  Find  the 
safe  working  pressure  in  pounds  per  square  inch.  Tensile  strength 
of  plate  is  56,000  Ibs.  per  sq.  in. 

PROB.  170.  A  return  tubular  boiler  is  84  ins.  in  diameter  and 
carries  a  pressure  of  125  Ibs.  per  sq.  in.  in  gauge.  The  longitudinal 
seam  is  a  triple-riveted  butt  joint.  Efficiency  of  joint  is  84  per  cent. 
Find  the  thickness  of  the  metal. 

PROB.  171.  (a)  In  the  above  problem  design  a  double-riveted 
lap  joint  for  the  girth  seam.  (6)  Which  is  the  weaker,  the  longitu- 
dinal or  the  girth  seam? 


CHAPTER  X 
COMBINED  STRESSES— RESILIENCE 

ART.  53. — CRANE  HOOKS — FLEXURE  AND  TENSION 

THERE  are  many  cases  of  problems  involving  a  combina- 
tion of  stresses.  The  mathematics  involved  in  some  of  these 
is  too  complicated  for  this  text,  so  the  student  is  referred  to 
higher  works  on  Applied  Mechanics.  There  are,  however, 
several  simple  cases.  One  of  these  is  the  combination  of 
flexure  or  bending  and  pure  tension  such  as  found  in  the 
design  of  crane  hooks. 

Fig.  91  shows  a  common  form  of  hook.  The  section  CD 
is  under  pure  tension,  and  the  unit  stress  on  this  section  is 

W 

found  from  the  formula  S  =  -r-,  where  W  is  the  load  in  pounds 

A. 

and  A  the  cross-section  in  square  inches  at  the  section  CD. 
The  area  EF  is  subjected  to  pure  shear,  hence  the  unit 

W 

shearing  stress  is  -p,  where  A\  is  the  area  at  section  EF. 
•Ai 

The  section  A B  is  subjected  to  a  pure  tensile  stress  due 
to  the  load  W  and  in  addition  is  subjected  to  a  bending  action 
due  to  the  fact  that  the  neutral  axis  of  the  section  A  B  is 
not  in  line  with  load  W.  This  bending  action  produces  a 
tension  in  the  part  to  the  right  of  the  neutral  axis  and  a  com- 
pression in  the  part  to  the  left  of  the  section.  Hence  the 
maximum  stress  is  in  the  outermost  fiber  at  the  point  B. 

Let  the  tensile  stress  due  to  the  pure  tension  equal 

151 


152 


STRENGTH  OF  MATERIALS 


W 


=  ^-,   and   let   the   tensile   stress   due   to    bending   equal 
b  =  -j-,  then  the  total  tensile  stress  on  the  outer  fiber  of 


the    section    AB    is 


FIG.  91. 


=     +?     The    bending 


FIG.  91a. 


moment  M  —  Wp,  where  p  is  the  distance  from  the  line  of 
action  of  the  weight  W  to  the  neutral  axis  of  the  section  AB. 
Hence  the  formula  for  the  design  of  the  section  AB  becomes, 


W    Wpc 


(100) 


where  S  equals  safe  unit  working  stress  in  tension,  W  equals 
the  load  in  pounds,  c  equals  the  distance  from  the  neutral  axis 


COMBINED  STRESSES—  RESILIENCE  153 

to  the  outermost  fiber  at  B,  and  I  equals  the  moment  of 
inertia  of  the  section  relative  to  the  axis  g-g. 

EXAMPLE.  Assume  the  hook  to  be  1J  ins.  in  diameter 
at  the  section  AB;  let  the  distance  p  =  2  ins.  The  hook  is 
made  of  wrought  iron;  let  £  =  8000  Ibs.  per  sq.  in.  Find  the 
safe  load  that  can  be  carried  by  the  hook. 

To  find  this  load  substitute  the  above  values  in  Equation 
(100)  thus: 

Q     W.Wpc 
fc-lH1—  ' 
where 

A  =  .  7854d2=1.77;  p  =  2ins., 

c=^.75;  1=^  =  0.25, 

hence, 


or 

W=  1200  Ibs. 

For  general  proportions  of  crane  hooks  the  student  is 
referred  to  handbooks  on  Machine  Design. 

For  a  section  of  the  form  shown  in  Fig.  91  it  is  necessary 
to  calculate  the  moment  of  inertia  about  the  g-g  axis.  This 
is  done  by  dividing  the  section  into  rectangles,  triangles, 
and  circular  sections,  and  apply  the  method  explained  in 
Art.  23. 

PROB.  172.  Find  the  safe  load  that  can  be  carried  by  a  hook  of 
the  form  shown  in  Fig.  91  if  the  depth  at  AB  is  2  ins.,  the  width  at 
point  A  is  &  in.  and  the  radius  at  B  is  ^  in. 

PROB.  173.  Find  the  unit  tensile  stress  in  the  section  CD,  if 
the  diameter  is  1|  ms- 

PROB.  174.  Find  the  unit  shearing  stress  in  the  section  EFt  if 
the  depth  of  the  section  is  1&  ins.,  width  at  narrow  end  is  A  in. 
and  radius  of  circular  section  at  E  is  ^  in. 


154 


STRENGTH  OF  MATERIALS 


ART.  54.     MACHINE  FRAMES 

The  stresses  set  up  in  machine  frames  are  usually  impos- 
sible to  analyze  mathematically,  owing  to  the  peculiar  nature 
of  the  external  load.  For  example,  the  frame  of  a  locomotive 
can  be  properly  designed  only  from  empirical  formulae  based 
on  actual  experience.  Again,  the  frame  of  a  machine  must 
be  rigid  enough  so  as  not  to  distort  any  of  the  work  being 
done  by  the  machine,  which  may  mean  the  use  of  very  high 


FIG.  92. 

factors  of  safety.  In  such  designs  experience  and  good  judg- 
ment are  often  more  essential  than  a  mere  analysis  of  the 
forces  acting.  There  are  cases  where  a  somewhat  clear 
analysis  of  the  various  stresses  can  be  made  to  good  advan- 
tage. 

Fig.  92  shows  an  open  type  of  frame  commonly  used  in 
the  construction  of  punch  presses  and  riveters.  These  frames 
are  generally  made  of  cast  iron  and  are  built  very  heavy. 
The  dangerous  section  of  the  frame  is  at  BC.  The  load  P 
will  cause  a  uniform  tension  on  the  section  BC.  Let  this 
tension  be  represented  by  S.  In  addition  there  will  be  a 
bending  action  on  the  section  BC  due  to  the  load  P  acting  at  a 
distance  of  a  units  from  the  neutral  axis  g-g.  Let  this  bend- 
ing moment  be  represented  by  PXa.  At  the  point  B  there 


COMBINED  STRESSES—  RESILIENCE  155 

will  be  a  unit  tensile  stress  due  to  the  bending  moment.  Let 
S  equal  this  stress.  Also  at  the  point  C  there  will  be  a  com- 
pressive  stress  due  to  the  bending  moment  PXa.  This 
cornpressive  stress  will  be  partially  counterbalanced  by  the 
direct  tensile  stress  due  to  the  load  P.  Since  cast  iron  is 
weaker  in  tension  than  in  compression  it  is  advisable  to  so 
design  the  section  that  the  neutral  axis  g-g  will  be  located 
near  the  point  B,  thus  putting  the  greater  area  of  the  section 
in  tension.  Let  S  equal  the  safe  unit  working  stress  in  ten- 
sion at  the  point  B.  It  is  evident  that  S  will  equal  the  sum 
of  St,  the  direct  tensile  stress,  and  Sb)  the  indirect  tensile 
stress  due  to  bending,  hence 


P  Me 

but  St  =  -7  and  S&  =  -^-,  where  M  equals  the  bending  moment 
A.  J- 

Pac 
PXa,  hence  S&  =  -j-. 

Then 

S  =  St~{-Sb  =  -7-H  —  j-j 
or 


In  Equation  (101)  c  is  the  distance  from  the  neutral 
axis  g-g  to  the  point  B,  and  /  is  the  moment  of  inertia  about 
the  axis  g—g.  The  value  of  r2  is  obtained  from  the  equation 

7*2  =  -j  ,  where  A  is  the  area  of  the  cross-section  in  square 

A. 

inches.     Figs.  46  and  51  show  various  forms  of  sections  used 
in  the  design  of  punch  presses. 

As  an  example  of  the  above,  consider  the  case  of  a  punch 
press  of  the  form  shown  in  Fig.  92  and  let  the  section  BC  be 
of  the  form  and  dimensions  given  in  Fig.  46.  The  16-in. 


156  STRENGTH  OF  MATERIALS 

width  of  the  section  will  be  placed  at  B.  For  this  section 
c=  12.65  and  /g-30407.  If  the  distance  a  =  30  ins.  let  it 
be  required  to  find  the  maximum  pressure  P  that  can  be 
exerted  by  the  press,  so  that  the  unit  tensile  stress  at  the 
point  B  shall  not  exceed  3000  Ibs.  per  sq.  in.  Solving 
Equation  (101)  for  P  gives 


where  A  =the  area  of  section  =  122  sq.  ins., 

c=  12.65, 
and 

^=1=250; 

hence 

P-  145,300. 

Let  d  equal  the  diameter  of  the  largest  hole  to  be  punched 
in  the  plate,  and  let  t  equal  the  thickness  of  the  plate  and  let 
S  equal  the  ultimate  shearing  strength  of  the  material,  then 
the  total  resistance  to  be  overcome  by  the  punch  is  irdXtXS; 
but  this  resistance  can  be  no  greater  than  the  force  P,  hence 

(102) 


From  Equation  (102)  the  maximum  diameter  of  hole  may  be 
figured  for  a  given  thickness  of  plate. 

Another  illustration  of  combined  bending  and  compression 
is  in  the  case  of  a  simple  overhanging  crane  of  the  form  shown 
in  Fig.  93.  The  unit  compressive  stress  at  the  section  mn 
is  equal  to  the  total  load  P,  divided  by  the  area  at  the  section 
inn.  In  addition  to  this  unit  compressive  stress  there  is  a 
bending  stress  due  to  the  bending  moment  exerted  by  the 


COMBINED  STRESSES— RESILIENCE 


157 


force  P.  This  bending  moment  is  equal  to  the  force  P  times 
the  swing  of  the  crane,  or  equals  PXa.  In  this  case  the 
maximum  stress  occurs  at  the  point  B  and  is  equal  to  the 
direct  unit  compressive  plus  the  indirect  compressive  stress 
due  to  the  bending  action.  Let  Sc  equal  the  direct  unit 

p 
compressive  stress  equal  -j  and  let  Sb  equal  the  indirect 


FIG.  93. 


Me 


compressive  stress  due  to  the  bending  moment,  then  *S>&  =    , 
where  M  =  PXa,  therefore 

P 


or 


(103) 


For  example,  let  the  section  at  mn  have  the  form  of  an 
/  section.  Let  the  width  at  point  n  equal  7  ins.,  at  the  point 
m  equal  5  ins.,  let  the  thickness  of  the  web  be  2  ins.  and  of  the 
flanges  3  ins.  Let  the  moment  arm  a  equal  24  ins.,  and 
depth  of  section  equal  10  ins.  If  the  unit  compressive  stress 
at  n  is  not  to  exceed  10,000  Ibs.  per  sq.  in.,  let  it  be  required 
to  find  the  maximum  load  P  that  can  be  supported  by  the 


158  STRENGTH  OF  MATERIALS 

crane.  The  first  step  in  such  a  problem  is  to  locate  the 
gravity  axis  of  the  section  and  then  determine  the  moment 
of  inertia  of  the  section  relative  to  the  neutral  axis  g-g. 
The  gravity  axis  is  located  by  taking  moments  about  the 
narrow  base  of  the  section.  Thus 

5X3X1.5+4X2X5+7X3X8.5=44z 

or. 

2  =  5. 5  ins.  (approx.) 

The  moment  of  inertia  about  the  g-g  axis  equals 

7X4. 53     5X1.53  .  5X5.53     3X2. 53_ 

"3~        ~3~        ~T~        ~T~ 
Also 

r2  =  |  =  ^  =  20  (approx.); 

hence  substituting  these  values  in  Equation  (103)  gives, 

10  000-P 
LO,OC    - 

or 

P  =  70,000  Ibs. 

Another  common  illustration  of  combined  bending  and 
tension  is  in  the  case  of  the  frame  of  a  side-crank  steam  engine. 
The  proper  frame  section  can  be  determined  by  an  analysis  of 
the  stresses  similar  to  that  in  the  case  of  the  punch-press 
frame.  Many  applications  of  this  principle  are  to  be  found 
in  structural  work. 

PROB.  175.  A  punch  press  having  a  gap  of  30  ins.  is  of  the  form 
and  dimension  shown  in  Fig.  51.  Find  the  maximum  size  hole  that 
can  be  punched  in  a  steel  plate  1|  ins.  thick,  so  that  the  maximum 
tensile  stress  on  the  dangerous  section  shall  not  exceed  2800  Ibs. 
per  sq.  in. 


COMBINED  STRESSES— RESILIENCE  159 


ART.  55.     MODULUS  OF  RESILIENCE 

In  Fig.  17,  which  shows  the  relation  between  the  load 
and  deformation  in  a  tension  test  it  will  be  noted  as  the  load 
is  applied  the  specimen  elongates,  and  hence  external  work 
is  done  upon  this  specimen.  If  the  load  is  removed  before 
the  elastic  limit  is  reached,  the  specimen  will  return  to  its 
original  length.  This  external  work  which  is  required  to 
stress  the  bar  is  called  "  resilience." 

Within  the  elastic  limit  the  deformation  increases  in  a 
direct  ratio  to  the  load  and  the  curve  from  no  load  to  the 
elastic  limit  becomes  a  straight  line.  If  the  load  at  the 
elastic  limit  is  P  pounds,  then  the  average  resistance  offered 

by  the  material  while  the  load  P  is  being  gradually  applied 

p 
is  —  pounds.     Let  d  equal  the  deformation  at  the  elastic 

limit.     It  is  apparent  that  the  work  done  on  the  specimen 

p 
equals  the  average  load  -^  times  the  deformation  d,  or  the 


,    Pd 
resilience  equals  -^-. 

The  resilience  may  be  represented  by  the  area  included 
between  the  y  axis  and  that  part  of  the  curve  which  extends 
froir  the  origin  to  the  elastic  limit.  This  area  takes  the  form 
of  a  triangle  whose  altitude  is  the  total  load  at  the  elastic 
limit  and  whose  base  is  the  total  deformation  at  the  elastic 
limit.  The  are.a  of  this  triangle  equals  one-half  the  base 

times  the  altitude,  or  equals  -^-,  which  is  the  same  as  the 

expression  for  the  resilience. 

Elastic  resilience  is  the  work  done  in  stressing  a  bar  to  the 
elastic  limit. 

Ultimate  resilience  is  the  work  done  in  rupturing  a  bar. 

Resilience  is  the  property  of  a  material  to  withstand 
external  work  being  done  upon  it.  The  total  elastic  resilience 


160  STRENGTH  OF  MATERIALS 

depends  upon  the  cross-sectional  area  and  length  of  a  given 
specimen.  In  order  to  compare  the  resilience  of  various 
materials  it  is  customary  to  state  the  resilience  in  terms  of  the 
work  done  upon  a  cubic  inch  of  the  specimen,  in  such  a  case 
the  work  done  upon  a  cubic  inch  of  the  material  is  called  the 
modulus  of  resilience.  For  example,  consider  the  case  of  a 
bar  of  steel  subjected  to  a  tensile  strain.  Let  A  equal  the 
original  cross-section  of  the  bar  in  square  inches,  L  equal  the 
length  in  inches,  d  equal  the  total  elongation  at  the  elastic 
limit,  P  equal  the  load  at  the  elastic  limit,  E  equal  the 
modulus  of  elasticity  of  the  material,  and  S  equal  the  unit 
stress  at  the  elastic  limit.  Then  by  definition 

„          unit  stress 


unit  deformation' 


and  the  elastic  resilience  equals   —  ^  —  .     The  unit  stress  at 

p 

the  elastic  limit  equals  S  =  -r,  or  P  =  AS.     Likewise  the  unit 

A. 

deformation  equals  y  and  hence 

Li 


E  = 


P 

unit  stress  A      PL 


unit  deformation      d      dA' 

L 
or 

,    PL 
=  AE' 

Let  K  equal  the  modulus  of  resilience,  then  substituting 
these  values  of  P  and  d  in  the  expression  for  elastic  resilience 
gives,  elastic  resilience  equals 

Pd  =  AS     PL  =       PL 

2  "  2      AE  2E' 


COMBINED  STRESSES—  RESILIENCE  161 

But  the  modulus  of  resilience  equals  the  total  resistance 
divided  by  the  volume  of  the  specimen  which  is  AL,  hence 


That  is,  the  modulus  of  resilience  of  any  material  is  equal 
to  the  square  of  the  unit  stress  at  the  elastic  limit  divided  by 
the  modulus  of  elasticity.  For  example,  the  elastic  limit  of 
steel  is  35,000  Ibs.  per  sq.  in.,  and  its  modulus  of  elasticity 
is  30,000,000,  hence  the  modulus  of  resilience  for  steel  equals 

350002 

—  ,  or  K  =  4l.     This  modulus  can  be  figured  for  com- 


30000000' 

pression  in  the  same  manner.  Equation  (104)  is  applicable 
only  to  either  a  tension  or  a  compression  test,  and  is  never 
to  be  used  in  connection  with  bending.  The  modulus  of 
resilience  is  expressed  in  inch-pounds  since  the  load  is  given 
in  pounds  and  the  deformation  in  inches.  For  timber  in 
tension  K  =  3  in.-lbs.;  for  cast  iron  K=l  in.-lb.,  and  for 
wrought  iron  K=  12  in.-lbs.  When  the  modulus  of  resilience 
is  known  the  work  required  to  stress  a  specimen  of  any  size 
can  be  figured;  for  example,  let  it  be  required  to  stretch 
a  steel  eye-bolt  which  is  2X4-ins.  in  cross-section  and  6  ft. 
long,  until  the  unit  stress  is  20,000  Ibs.  per  sq.  in.  Assume 
that  the  load  is  to  be  applied  in  thirty  seconds,  find  the 
horse-power  required.  The  total  load  at  the  point  of  maxi- 
mum stress  is  20,000  X  2  X4  =  160,000  Ibs.  The  unit  deforma- 

S       20000  1 

tion  due  to  the  unit  stress  of  20,000  Ibs.  is  -^  =  ; 


E    30000000     1500 

1         72 
ins.     The  total  deformation  is  72X^.^  =  3-^.^  ins.,    since 


160000 
the  bar  is  72  ins.  long.     The  average  load  exerted  is  - 

A 

72 
=  80,000  Ibs.,  hence  the  total  work  done  equals  80,000  X^ 


1500 


162  STRENGTH  OF  MATERIALS 

=  3840  in.-lbs.     The  work  done  per  minute  is  3840 X  2  =  7680 
in.-lbs.     Hence  the  horse-power  required  to  stress  the  bar  is 

7689 
12X33000  =  -° 

PROB.  176.  Figure  the  elastic  resilience  of  a  wrought-iron  tie 
rod  which  is  2x3.5  ins.  in  cross-section  and  10  ft.  long  when  sub- 
jected to  a  unit  stress  of  15,000  Ibs.  per  sq.  in. 

PROB.  177.  How  could  the  ultimate  resilience  be  figured  in 
the  case  of  a  steel  specimen  under  tension? 


ART.  56.     FORMULA  FOR  ELASTIC  RESILIENCE 

In  the  case  of  beams  the  elastic  resilience  is  the  product 
of  the  deflection  and  the  total  load  applied  provided  the  load 
is  such  as  to  produced  unit  stress  in  the  beam  not  exceeding 
the  elastic  limit.  If  P  is  the  load  and  d  the  deflection,  then 

the  resilience  or  work  done  on  the  specimen  is  -^-,  as  the 

2i 

load  varies  from  0  to  P.  The  modulus  of  resilience  is  equal 
to  the  elastic  resilience  divided  by  the  volume  of  the  beam, 
which  is  A  XL,  where  A  is  the  cross-sectional  area  and  L  is 
the  length  of  the  span  in  inches.  Let  K  equal  the  modulus 
of  resilience,  then, 


which  is  a  general  expression  for  the  elastic  resilience  of  beams. 

Several  special  cases  will  be  considered. 

CASE  I.     Simple  beam  of  span  L  carrying  a  concentrated 
load  of  P  pounds  at  the  center  of  the  span.     In  this  case  the 

1  PL3 

deflection  equals  -.  •  -j^j-.     The  unit  stress  due  to  the  load 
4o  rLl 

Me  PL 

P  is  $=-jr-,  but  for  a  concentrated  load  M  =  —  ;    hence 


COMBINED  STRESSES—  RESILIENCE  163 

£  =  -TT->  or  F  =  ~]~'     Substituting  this  value  of  P  in  the 
expression  for  deflection  gives 


_         =  = 

°  "48  El     48  Lc     El     l2Ec 

To  express  the  modulus  of  resilience  in  terms  of  the  unit 
stress  substitute  the  values  of  P  and  d  in  the  Equation, 

Pd        ,  .  , 

-j  which  gives 

481      SL2          1         *S2r2 

PC  —  —  —  V—  —  -  -V—  Hftftt 

"  LcXl2EcX2AL~GEc2' 

In  Equation  (106)  I  has  been  replaced  by  its  equivalent  value 
I  =  Ar2. 

If  the  total  resilience  is  desired,  it  may  be  found  by  cal- 
culating the  modulus  of  resilience  from  Equation  (106)  and 
multiplying  this  value  by  the  volume  of  the  specimen. 

CASE  II.  Simple  beam  of  span  L  carrying  a  uniformly 
distributed  load  of  W  pounds  per  linear  foot.  In  this  case 
the  modulus  of  resilience  is  found  from  Equation  (105)  by 
substituting  for  P  the  total  uniform  load  W  and  expressing 
W  in  terms  of  the  unit  stress.  For  a  simple  beam  uniformly 

.  ,,    WL        ,  „    Me    WLc        w    SSI      ...      . 
loaded  M  =  —^-,  and  o  =  -y-  =-=-=-,  or  W  =  ^~.     Likewise 

o  .     JL  ol  L/C 

5 


2AL     Lc 
or 


The  ultimate  resilience  of  a  beam  cannot  be  calculated 
from  any  formula,   owing  to   the  variable  nature  of  the 


164  STRENGTH  OF  MATERIALS 

relations  between  load  and  deformation.  Furthermore,  the 
ultimate  resilience  has  little  practical  value  in  the  design  of 
structures  and  machines. 

PROB.  178.  Compute  the  modulus  of  resilience  of  a  simple 
yellow-pine  beam  of  12  ft.  span  carrying  a  concentrated  load, 
such  that  the  unit  fiber  stress  is  equal  to  the  elastic  limit  of  the 
material.  The  beam  is  6x9  in  cross-section. 

PROB.  179.  Compute  the  modulus  of  resilience  of  a  light  10-in. 
steel  I  beam  of  24-ft.  span.  Beam  is  uniformly  loaded,  so  that  unit 
fiber  stress  equals  the  elastic  limit  of  steel. 


ART.  57.     STRESSES  DUE  TO  TEMPERATURE 

When  a  steel  or  iron  member  is  subjected  to  a  change  of 
temperature  there  is  a  corresponding  decrease  or  increase  in 
the  length  of  the  member,  depending  upon  whether  the  change 
in  temperature  is  a  decrease  or  an  increase.  If  the  member 
is  free  to  move  there  will  be  no  strain  put  upon  the  member, 
but  if  the  member  is  fixed  so  that  it  cannot  shorten  or 
lengthen,  then  a  unit  stress  will  be  set  up  which  is  the*  same 
as  an  equivalent  unit  stress  required  to  produce  the  same 
change  of  length  as  that  due  to  the  change  of  temperature. 
If  there  is  an  appreciable  change  of  temperature  and  the 
length  of  the  structure  is  large,  then  provision  must  be  made 
by  suitable  expansion  joints  to  allow  for  the  change  of  length. 
If  the  structure  is  rigid,  the  stress  due  to  temperature  may 
be  great  enough  to  cause  "  buckling  "  or  collapse  of  the 
structure. 

The  change  in  length  depends  upon  the  material  and  upon 
the  change  of  temperature.  The  "  coefficient  of  linear  ex- 
pansion "  of  a  given  material  is  the  elongation  per  unit  of 
length  per  degree  difference  of  temperature.  Table  XVIII 
gives  the  coefficient  of  expansion  for  various  metals  based 
on  a  change  of  temperature  of  1°  F: 


COMBINED  STRESSES— RESILIENCE 


165 


TABLE   XVIII 


Material. 

Coefficient  of 
Expansion. 

/  Hard  
1  Structural  

.0000074 
.0000061 

Cast  iron  

.0000063 

Wrought  iron  

.0000068 

Stone  and  brick  

.0000050 

Copper  

.0000094 

Concrete 

.  0000055 

Let  C  equal  the  coefficient  of  expansion,  t  equal  the  change 
of  temperature  in  degrees  Fahrenheit,  and  E  equal  the 
modulus  of  elasticity.  The  unit  stress  due  to  a  given  elon- 
gation is  E  multiplied  by  the  elongation,  as  the  modulus  of 
elasticity  is  equal  to  the  unit  stress  divided  by  the  unit 
elongation.  But  the  unit  elongation  for  a  change  of  t  degrees 
is  Ct,  hence  the  unit  stress  equals 


=  CtE. 


(108) 


It  will  be  noted  that  this  unit  stress  is  independent  of  the 
length  of  the  specimen.  If  the  specimen  is  free  to  change  in 
length,  the  total  elongation  equals  the  unit  elongation  times 
the  length  of  the  specimen.  For  example,  assume  the  case  of 
a  steel  I  beam  40  ft.  long.,  which  is  free  to  move  at  the  sup- 
ports. Let  the  change  of  temperature  be  an  increase  of 
70°  F.,  what  will  be  the  increase  in  length  of  the  beam? 
From  Table  XVIII,  1°  F.  will  produce  a  change  in  length  of 
.0000061  per  unit  of  length,  that  is,  if  the  beam  is  1  ft.  in 
length,  for  each  degree  increase  of  temperature  the  beam  will 
increase  .0000061  ft.  in  length;  hence,  for  a  beam  40  ft.  long 
and  a  change  of  temperature  of  70°  F.  the  increase  in  the 
length  of  the  beam  is  .0000061X40X70  =  .0171  ft.  =  2.05  ins. 
If  the  beam  decreases  in  temperature,  then  the  beam  will  be 
shortened  by  an  amount  equal  to  the  above. 


166  STRENGTH  OF  MATERIALS 

As  another  example,  consider  the  case  of  a  steel  beam 
which  is  supported  between  two  rigid  columns.  Assume  the 
length  of  beam  to  be  50  ft.  under  normal  temperature.  If 
the  temperature  drops  50°  F.,  a  tensile  stress  will  be  set  up  in 
beam,  while,  if  the  temperature  increases  the  beam  will 
be  in  compression.  The  value  of  this  unit  stress  is  given  by 
Equation  (108),  thus 

S  =  CtE  =  .000006 1 X  50  X  30,000,000 
=  9150  Ibs.  per  sq.  in. 

In  designing  structures  proper  allowance  must  be  made 
for  the  probable  change  in  temperature,  for  example,  if  steel 
rails  were  laid  tight  up  against  one  another,  when  a  change  of 
temperature  occurred  the  internal  stress  would  be  sufficient 
to  cause  a  deformation  of  the  rails. 

Another  common  illustration  of  stress  due  to  change  ot 
temperature  is  in  the  case  of  "  shrink  "  fits.  For  example, 
the  steel  tires  on  the  driving  wheels  of  a  locomotive  are  held 
in  place  by  shrink  fits,  that  is,  the  tire  is  turned  to  an  inside 
diameter  somewhat  less  than  the  outside  diameter  pf  the 
wheel.  The  tire  is  then  heated  and,  of  course,  increases  in 
diameter.  After  the  proper  diameter  is  secured  the  tire  is 
slipped  over  the  wheel  and  allowed  to  cool,  thus  contracting 
and  firmly  gripping  the  wheel.  The  initial  diameter  must  be 
such  that  when  the  tire  cools  the  unit  stress  due  to  the  con- 
traction shall  not  exceed  a  predetermined  amount.  If  the 
diameter  of  the  wheel  is  large,  and  the  thickness  of  the  tire 
very  small  compared  to  the  wheel,  then  the  diameter  of  the 
wheel  remains  practically  constant.  In  such  a  case  let  D 
equal  the  diameter  of  the  wheel  and  d  equal  the  inside  diam- 
eter to  which  the  tire  is  turned.  Then  the  unit  change  in 

the  length  of  the  tire  when  placed  on  the  wheel  is  -  — ^ — -, 
or    this   quantity    represents   the    unit    deformation.     The 


COMBINED  STRESSES—  RESILIENCE  167 

corresponding  unit  stress  is  S  =  E  times  the  unit  deformation 
equals 

=.        .....     (109) 


If  the  stress  is  sufficient  to  cause  a  decrease  in  the  diameter 
of  the  wheel,  then  the  above  formula  is  not  applicable. 
This  case  is  beyond  the  scope  of  this  text.  Large  guns  are 
constructed  by  shrinking  a  series  of  jackets  onto  the  barrel 
of  the  gun.  As  an  example,  consider  the  case  of  a  cast-iron 

driving  wheel  which  is  turned  to  a  diameter  ^-=7^:  greater 

luUU 

than  the  diameter  of  the  tire.     Here  the  unit  deformation  is 

x  and  the  unit  stress  is  S  =  EXTl  ^  =  20,000  Ibs. 
a         loUU  15UU 

per  sq.  in. 

PEOB.  180.  A  wrought-iron  tie  rod  holds  together  the  two 
heads  of  a  boiler.  Find  the  unit  stress  in  the  tie  rod  when  the 
temperature  of  the  rod  is  changed  400°  F.  Assume  that  the  heads 
of  the  boiler  remain  rigid. 

PROB.  181.  What  will  be  the  effect  of  a  change  in  tempera- 
ture on  a  beam  which  is  made  up  of  part  concrete  and  part  steel? 


ART.  58.     REVIEW  PROBLEMS 

PEOB.  182.  In  a  W.  I.  crank  hook  of  the  form  shown  in  Fig.  91 
the  distance  AB  is  5  ins.,  width  at  the  point  A  is  1  in.  and  the  radius 
at  B  is  1  in.  The  distance  p  is  8  ins.  Find  the  maximum  load  that 
can  be  placed  at  W  so  that  the  greatest  unit  stress  in  tension  shall 
not  exceed  7000  Ibs.  per  sq.  in. 

PEOB.  183.  In  the  above  hook  determine  the  diameter  at  CD 
(see  Fig.  91),  allowing  a  factor  of  safety  of  six. 

PEOB.  184.  In  the  above  problem  select  dimensions  for  the 
section  EF,  so  that  the  unit  shear  shall  not  exceed  6000  Ibs.  per  sq.  in. 

PROB.  185.  In  a  punch  press  of  the  form  shown  in  Fig.  92  and 
of  the  cross-section  shown  in  Fig.  51,  determine  the  distance  a  if  an 


168  STRENGTH  OF  MATERIALS 

inch  hole  is  to  be  punched  in  a  plate  £  in.  thick,  assuming  the  ulti- 
mate shearing  strength  of  the  plate  as  52,000  Ibs.  per  sq.  in. 

PROB.  186.  Figure  the  elastic  resilience  when  a  boiler  plate 
specimen  18  ins.  long  and  f  X3  ins.  in  section  is  subjected  to  a  unit 
tensile  stress  of  18,000  Ibs.  per  sq.  in. 

PROB.  187.  Derive  an  expression  for  the  elastic  resilience  of  a 
cantilever  beam  with  a  uniformly  distributed  load. 

PROB.  188.  Assuming  the  same  cross-section  and  span  com- 
pare the  relative  resilience  of  the  four  common  types  of  beams. 

PROB.  189.  Steel  rails  are  laid  in  place  at  a  normal  tempera- 
ture of  50°  F.  What  must  be  the  clearance  between  the  ends  of  the 
rails  to  allow  an  increase  of  temperature  to  110°  F.?  Assume  the 
rails  to  be  30  ft.  in  length. 

PROB.  190.  A  wrought-iron  steam  line  is  500  ft.  in  length. 
If  the  pipe  is  set  in  place  at  70°  F.,  what  will  be  the  increase  in  the 
length  of  the  pipe  when  the  temperature  is  raised  to  400°  F. 


CHAPTER  XI 
REINFORCED  CONCRETE 

ART.  59.     CEMENT  AND  CONCRETE 

CONCRETE,  which  is  one  of  the  most  widely  used  materials 
of  construction,  is  an  artificial  stone,  made  by  mixing,  in 
various  proportions,  Portland  cement,  sand,  and  crushed 
stone,  gravel,  or  cinders.  Since  it  is  really  stone,  it  has  its 
greatest  value  in  resisting  compressive  stresses. 

Portland  cement,  which  is  the  building  factor  in  con- 
crete, received  the  name  Portland  because  one  of  the  earliest 
English  makers  of  this  product  noticed  that  when  it  hard- 
ened it  very  closely  resembled  the  stone  found  in  the  quarries 
on  the  island  of  Portland,  off  the  south  coast  of  England. 
It  is  not  a  trade  name,  and  should  not  be  confused  as  such. 
In  this  country  to-day  we  have  a  great  many  concerns  all 
making  Portland  cement,  but  each  company  uses  its  own 
name,  as  for  example,  Atlas,  Alpha,  Lehigh,  Edison,  Universal 
and  so  on.  It  is  made  by  grinding  very  finely  the  clinker 
which  results  from  burning  at  a  high  temperature  a  mixture 
of  two  finely  ground  stones,  one  containing  a  high  percentage 
of  lime  and  the  other  a  high  percentage  of  alumina  and  silica. 
These  are  the  three  principal  elements  in  Portland  cement, 
and  it  is  necessary  to  use  two  different  kinds  of  stone  in  order 
to  obtain  the  proper  amounts  of  each.  It  is  this  scientific 
mixing  of  the  powdered  stones  which  really  makes  Portland 
cement  different  from  other  cements. 

169 


170  STRENGTH  OF  MATERIALS 

Portland  cement  is  mixed  with  sand  to  make  mortar,  and 
with  sand  and  crushed  stone  to  make  concrete.  The  sand 
and  stone  are  frequently  referred  to  as  aggregates,  fine  and 
coarse.  Sand  is  the  name  given  to  the  material  which  will 
pass  through  a  quarter-inch  screen,  and  the  remaining  mate- 
rial which  is  too  large  to  pass  through  the  quarter  screen  is 
called  stone  or  gravel.  The  principal  requisite  of  these  two 
materials  is  that  they  be  hard,  durable,  clean,  and  free  from 
loam  or  other  vegetable  matter.  It  is  not  necessary  that  the 
particles  of  the  aggregates  be  pointed  and  rough,  although 
this  is  sometimes  helpful.  The  size  of  the  largest  piece  of 
stone  to  be  used  in  making  concrete  depends  on  the  purpose 
for  which  the  concrete  is  made.  They  usually  are  of  a  size 
that  will  pass  through  a  f-in.  or  1-in.  diameter  ring  for  rein- 
forced slabs  and  beams,  and  may  be  of  2  ins.  or  2|  ins.  size 
for  large  masses  such  as  engine  foundations,  columns,  foot- 
ings, retaining  walls,  etc.  In  any  event,  the  pieces  should 
be  cubical  in  shape  rather  than  flat  or  elongated ;  and  should 
be  graded  in  size  from  the  largest  down  to  the  quarter-inch 
pieces. 

The  cement  sand  and  stone  are  mixed  together  in  various 
proportions  by  volume.  It  is  customary  to  express  these 
proportions  numerically  thus :  1  :  2  :  4,  1  :  3  :  6,  which  means 
one  part  by  volume  of  cement,  two  parts  sand,  and  four  parts 
stone.  The  word  "  part  "  may  mean  cubic  foot,  cubic  yard, 
or  even  cubic  inch.  Since  a  bag  of  cement  holds  approxi- 
mately 1  cu.  ft.  it  is  convenient  to  use  the  cubic  foot  as  a 
basis  of  measurement.  The  following  table  will  be  found 
helpful  in  estimating  the  quantities  of  materials  required  to 
make  a  given  amount  of  mortar  or  concrete: 


REINFORCED  CONCRETE 
TABLE  XIX 


171 


Mix. 

QUANTITIES  O] 

-  MATERIALS  IN 
MORTAR. 

1  Cu.   FT.  OF 

Cement,  Bag. 

Sand,  Cu.  Ft. 

1  :  I-*  

.592 

.840 

1:2       

.496 

.919 

1  :  2| 

421 

.999 

1  :  3  

.367 

1.053 

QUANTITIES  OP 

MATERIALS  IN 
CONCRETE.     • 

1   Cu.   FT.  OF 

Cement,  Bag. 

Sand,  Cu.  Ft. 

Stone  or  Gravel, 
Cu.  Ft. 

12:4 

232 

.440 

.880 

1     2|  :  5  

.192 

.459 

.921 

1     3    :  6  

.161 

.470 

.940 

It  is  customary  to  use  the  1:2:4  mix  for  reinforced  con- 
crete and  wherever  a  dense,  smooth  surface  or  water-tight 
concrete  is  desired.  If  compressive  strength  alone  is  all 
that  is  required,  a  1  :  2J  :  5  mix  will  probably  do  and  for 
large  masses  such  as  engine  beds,  foundations,  etc.,  riot  sub- 
ject to  vibration  or  shock,  a  1  :  3  :  6  can  be  used.  To  illus- 
trate how  Table  XIX  is  used,  suppose  we  wanted  to  make  50 
cu.  ft.  of  mortar  of  1  :  2J  mix,  and  125  cu.  ft.  of  concrete  of 
1  :  2J  :  5  mix.  We  would  need  for  the  mortar: 

SOX. 421  =  20.15,  say  20  bags  of  cement; 
50 X. 999  =49.45,  say  50  cu.  ft.  of  sand; 

NOTE. — 50  cu.  ft.  of  sand  equals  1.9,  say  2  cu.  yds.; 
and  for  the  concrete 

125 X .192  =  240  bags  of  cement; 

125 X. 459  =  57.4,  say  58  cu.  ft.  of  sand; 

125X.921=  115.1,  say  115  cu.  ft.  of  stone; 


172  STRENGTH  OF  MATERIALS 

NOTE. — 57.4  cu.  ft.  =  2.12,  say  2|  cu.  yds  of  sand; 
and  115  cu.  ft.  =4.26,  say  4j  cu.  yds  of  stone 

PROB.  191.  How  many  bags  of  cement  and  how  many  cubic 
feet  of  sand  are  required  to  make  the  mortar  1  :  3  mix,  for  a  brick 
pier  8  ft.  high  by  32  ins.  square,  if  there  are  340  cu.  ins.  of 
mortar  in  each  cubic  foot  of  brickwork. 

PROB.  192.  How  many  barrels  of  cement,  cubic  feet  of  sand 
and  cubic  feet  of  stone  will  be  required  to  build  an  engine  founda- 
tion 10  ft.  long  by  6  ft.  wide  and  4  ft.  high  of  stone  concrete 
1  :  2^  :  5  mix.  (NOTE. — There  are  four  bags  of  cement  in  one 
barrel.) 

ART.  60.     TYPE  OF  BARS 

Concrete  as  a  material  used  for  building  construction  has 
strength  only  in  compression.  However,  the  fact  that  con- 
crete, when  hardening,  takes  a  firm  grip  on  any  steel  rods 
which  may  be  embedded  in  it  enables  us  to  use  concrete, 
when  so  reinforced  by  steel  rods,  in  places  where  it  is  sub- 
jected to  tensile  as  well  as  compress! ve  stresses.  The  steel 
rods  take  up  the  tension  and  transfer  it  to  the  conrcrete. 
In  this  way  we  can  successfully  build  floor  slabs,  beams,  and 
columns  of  reinforced  concrete.  It  is  possible  to  reinforce 
concrete  with  plain,  round,  or  square  bars,  and  buildings 
which  have  been  so  reinforced  have  withstood  all  stresses 
and  vibrations.  It  is,  however,  a  fact  that  some  type  of 
"  deformed  "  bar  will  have  a  better  bond  with  the  concrete 
and  require  a  greater  force  to  pull  it  out.  For  this  reason 
most  engineers  prefer  to  use  the  deformed  bars  since  it  is 
possible  to  use  a  smaller  number  of  them,  in  other  words,  a 
square  inch  of  steel  in  a  deformed  bar  will  stand  a  larger  pull 
than  in  a  plain  bar.  There  are  a  great  many  types  of  de- 
formed bars  now  being  used,  and  it  is  impossible  to  illustrate 
them  all  here.  A  few  of  the  more  commonly  used  forms  are 
shown  in  Fig.  94  at  a,  6,  c,  and  d.  The  bar  shown  at  a  is 


REINFORCED  CONCRETE 


173 


probably  the  most  commonly  used  of  all  the  deformed  bars. 
There  is  no  patent  on  this  bar,  and  the  additional  cost  of  twist- 


a.  Twisted  bar. 


6.  Corrugated  square  bar. 


:.  Havemeyei  bar. 


d.  Triangle  Mesh  Reinforcement. 

FIG.  94. 


ing  is  not  very  great.  The  bars  shown  at  b  and  c  are  made 
in  both  square  and  round  sections,  and  at  d  is  shown  one  of 
the  common  types  of  reinforcement  used  in  floor  slabs. 


174 


STRENGTH  OF  MATERIALS 


ART.  61.     SLABS  AND  BEAMS 

Reinforced  concrete  construction  can  be  divided  into 
three  general  classes:  Floor  slabs,  beams,  and  columns. 
Floor  slabs  are  generally  thin  (about  4  ins.  being  an  average 
thickness),  and  their  span  is  usually  limited  to  from  6  to  8  ft., 
although  larger  spans  may  be  used.  The  slab  may  be  sup- 
ported by  steel  beams,  as  shown  in  Fig.  95,  or  may  be  cast 


FIG.  95. 

as  part  of  the  whole  floor  system,  as  indicated  in  Fig.  96. 
Fig.  95  shows  the  usual  arrangement  of  concrete  and  steel  in  a 
slab.  Notice  that  the  reinforcement  (which  is  here  indicated 
as  mesh,  but  the  arrangement  would  be  the  same  for  rods) 
is  not  straight  through  the  slab,  but  is  near  the  lower  surface 
or  near  the  middle  of  the  span  and  near  the  upper  surface 


where  it  passes  over  the  beams.  Care  must  be  taken  to  see 
that  this  arrangement  is  maintained.  The  beams  are  the 
members  which  support  the  slabs  and  are,  in  turn,  supported 
by  the  girders,  which  carry  the  load  to  the  columns  or  walls. 
The  beams  and  girders  are  reinforced  by  placing  the  rods 


REINFORCED  CONCRETE  175 

near  the  lower  surface,  but  since  the  stresses  for  which  the 
rods  are  provided  decrease  toward  the  ends  of  the  beams,  it  is 
possible  to  bend  up  some  of  the  rods  diagonally  across  the 
beams  and  in  this  way  help  to  provide  reinforcement  against 
shearing  stresses.  The  way  in  which  the  rods  are  bent  is 
shown  in  Fig.  97.  When  the  beams  or  girders  form  part  of 


\      \        \            \ 

x'  X_  /__/_  

FIG.  97. 

a  floor  system  it  is  customary  to  bend  the  roads  again  and 
carry  them  over  into  the  adjoining  beam  or  girder,  thus 
causing  the  beams  to  act  as  continuous  beams. 

ART.  62.     COLUMNS 

Concrete  columns  can  be  built  plain,  that  is,  without  any 
reinforcement,  but  if  they  are  reinforced  they  can  be  made 
of  smaller  sections,  since  the  effect  of  the  reinforcement  is 
to  increase  the  allowable  unit  stress  in  the  concrete.  Col- 
umns may  be  reinforced  in  a  number  of  ways.  Fig.  98 
shows  three  column  sections,  each  of  which  is  reinforced 
with  vertical  rods  only.  At  a  is  shown  a  square  column 
with  eight  reinforcing  rods,  at  b  an  octagonal  column  with 
eight  rods,  and  at  c  a  circular  column  with  sixteen  rods. 
Fig.  99  shows  two  columns,  the  first  with  "  hoop  "  reinforcing 
(the  hoops  must  be  placed  as  the  column  is  built),  and  the 
second  with  "  spiral  "  reinforcing.  Either  of  these  two 
methods  is  effective,  but  not  frequently  used  on  account  of 
difficulties  in  construction.  The  most  effective  and  fre- 
quently used  method  of  reinforcing  columns  is  shown  in 
Fig.  100.  Here  we  have  a  combination  of  the  two  methods, 


176 


STRENGTH  OF  MATERIALS 


FIG.  98. 


"Hoop"  "Spiral" 

FIG.  99. 


FIG.  100, 


REINFORCED  CONCRETE  177 

and  the  vertical  rods  are  combined  with  hoops  or  spirals. 
This  reinforcement  can  be  used  for  square,  octagonal,  or  cir- 
cular sections,  and  any  desired  number  of  rods  a  can  be 
placed  inside  the  spiral  6.  The  rods  and  spiral  should  be 
securely  fastened  together  at  frequent  intervals,  so  that  the 
spacing  shall  be  maintained  during  the  pouring  of  the  con- 
crete. 

ART.  63.     DESIGN  OF  SLABS  AND  BEAMS 

Since  concrete  has  very  little  strength  in  tension,  it  be- 
comes necessary  to  place  steel  rods  in  the  lower  part  of  con- 
crete beams  to  take  care  of  the  tensile  stresses  which  exist 
there.  It  should  be  kept  in  mind  that  this  is  made  possible 
because,  as  has  been  previously  stated,  the  concrete  takes  a 
firm  grip  on  the  steel  when  it  sets  or  hardens.  The  stresses 
which  exist  in  a  reinforced  concrete  beam  are  the  same  as 
those  in  any  other  beam  and,  therefore,  will  not  be  discussed 
here.  There  are  a  number  of  terms  and  formulae  which  are 
peculiar  to  reinforced  concrete  design  and  these  will  be  stated 
and  discussed  briefly,  before  illustrating  their  use  by  means 
of  numerical  examples.  The  following  notation  will  be  used 
in  the  formulae: 

Sc  =  Maximum  unit  stress  in  the  concrete  caused  by  a 

given  load; 
$s  =  Unit  stress  in  the  steel  caused  by  the  same  load; 

(7  =  Total  compression  in  the  concrete; 

T  =  Total  tension  in  the  steel; 
Es  =  Modulus  of  elasticity  of  steel; 
Ec  =  Modulus  of  elasticity  of  concrete; 

n  —  Ratio  — ; 

d  =  Distance  from  top  of  concrete  to  axis  of  steel  (not 

total  depth  of  beam) ; 
6  =  Breadth  of  rectangular  beam ; 


178 


STRENGTH  OF  MATERIALS 


As  =  Area  of  steel; 

As 
p  =  Steel  ration  =  —  or  A  s  =  pbd  ; 

M  =  Maximum  bending  moment  (in  inch-pounds)  due  to 
given  load. 

Both  k  and  j  are  coefficients  which  may  be  considered  as 
follows:    If  the  distance  from  the  top  of  the  beam  to  the 

I  I 6— >i 


t 

=^    gleo 

1 

Neutral  Axis 

3 

-» 

.<Area  of  Steel 
=  As 

Stress  Diagram          ^?  Cross  Section  at  B-B 

FIG.  101. 
neutral  axis  were  called  x  then  the  value  of  k  is  such  that 

k  =  -  or  x  =  kd,  and  it  is  so  shown  in  Fig.  101.     In  a  similar 
a 

manner  the  distance  from  the  center  of  compression  to  the 
steel  axis  is  represented  by  jd. 

Many  of  the  formulae  in  reinforced  concrete  work  seem 
long  and  complicated,  but  it  is  not  as  difficult  ^o  use  them  as 
it  seems. 

The  neutral  axis  is  first  located  by  the  formula 


k  = 


2  — 


pn. 


(110) 


Then  j  is  found  by  remembering  that  the  center  of  com- 
pression is  at  the  center  of  gravity  of  the  triangle  representing 
the  stress  in  the  concrete. 


7    kd  .          k 

=  d-         or      =  1-- 


(Ill) 


REINFORCED  CONCRETE 


179 


The  resisting  moment  of  the  beam  can  be  found  by  taking 
moments  about  the  center  of  compression  or  the  steel  axis 
instead  of  about  the  neutral  axis,  as  was  done  in  the  case  of 
ordinary  beams.  This  is  due  to  the  fact  that  all  the  tension 
passes  through  the  steel  instead  of  being  distributed  over  the 
lower  part  of  the  section  of  the  beam.  Since  the  resisting 
moment  is  equal  to  the  bending  moment  we  can  say  either 


or 


=  CXjd, 
=  TXjd. 


&(?£#•!& 

Neutral  
-Steel 

:f3p$t 
$%f&% 

—  9| 

! 

W  1« 



Axis 

i^Ss,*- 

^&— 

FIG.  102. 
From  Fig.  102  we  see  that  C,  the  total  compression  in  the 

concrete,  is  equal  to  kdX-^-Xb  and  T  the  total  tension  in  the 

fi 

steel  is  equal  to  ASXSS.    Therefore  we  have 


(112) 


or 


=AsSsXjd.     ......     (113) 

From  Equation  (112)  we  obtain 

2M 


and  from  Equation  (113) 


Since  As  =  pbd. 


bkjd2' 


M        M 


Asjd    pbjd2' 


(114) 


(115) 


180  STRENGTH  OF  MATERIALS 

A  simple  beam  of  reinforced  concrete  could  not  be  designed 
by  finding  .the  value  of  M  due  to  the  load  and  span,  and  then 
assuming  values  for  p,  n,  and  d.  k  would  be  found  by  Formula 
(110)  andj,  by  Formula  (111).  Then,  by  assuming  a  value 
for  Ss  the  proper  value  of  6  would  be  found  by  Formula  (115), 
for  which 

,        M 

0   =  ~ 


Finally  Sc  would  be  found  by  Formula  (114).  This  value  of 
Sc  must  be  within  the  safe  limits  of  concrete  in  compression 
or  else  the  whole  calculation  must  be  done  over  again  taking 
a  new  value  for  p.  It  is  not  usual  practice  to  make  as  many 
assumptions  as  would  be  indicated  by  the  foregoing,  but  this 
has  been  stated  to  indicate  what  the  procedure  would  be  like 
and  also  to  call  attention  to  the  fact  that  the  various  values 
of  p,  n,  and  Ss  are  not  absolutely  fixed,  but  may  vary  for  dif- 
ferent conditions. 

Since  floor  slabs  are  the  most  usual  form  of  rectangular 
beam  in  general  construction,  one  will  be  used  here  for  dis- 
cussion. A  beam  is  most  economical  when  it  is  so  designed 
that  the  maximum  allowable  unit  stress  in  the  steel  is  realized 
under  the  same  load  that  causes  the  maximum  allowable  unit 
stress  in  the  concrete.  From  this  it  can  be  seen  that  there  is  a 
certain  amount  of  steel  which  will  combine  with  a  given  area 
of  concrete  to  cause  a  given  unit  stress  in  the  steel  to  exist  at 
the  same  time  that  a  given  unit  stress  exists  in  the  concrete. 
This  amount  of  steel  can  be  indicated  by  p  and  it  can  be 
found  by  the  formula 


s 

2    2«/._°L 
&\178, 

Since  the  value  of  p  depends  only  on  Ss,  Se,  and  n,  it  is  pos- 
sible to  find  values  of  p,  k,  and  j  which  can  be  used  for  all 
rectangular  beams.  Thus,  it  is  recognized  as  good  prac- 


REINFORCED  CONCRETE  181 

tice  to  use  16,000  for  Ss,  650  for  Sc,  and    15  for  n  (Ea  = 
30,000,000  and  Ec  =  2,000,000). 


Solving  Formula  (116)  for  p  we  get 


=-00769- 


"* 


650     15X650; 
From  Formula  (110)  we  get 

k  =  V2  X  .00769  X  15  +  (.00769  X  15)2  -  .00769  X  15  =  .379 
and  from  Formula  (111)  we  get 


j  =  l-     -=1-.126  =  . 

o 

From  Formula?  (114)  and  (115)  we  can  get 

,, 

2      ' 
or 

Sspbjd2 

and  by  substituting  in  these  the  values  of  p,  k,  and  j  just 

found,  we  find  that 

M=W7Abd2. 

It  is  customary  to  bend  -the  alternate  rods  in  the  slab  up 
into  the  upper  part  at  the  ends  near  the  beams  and  to  con- 
tinue them  over  the  beam  near  the  top  surface.  For  this 
reason  it  is  the  usual  practice  to  consider  the  value  of  M  as 

—  —  where  W  =  total  load  and  L  =  the  span  in  inches. 

A  portion  of  the  slab  1  ft.  wide  is  considered  as  a  beam 
and  the  depth  and  area  of  steel  figured  for  this.  Thus,  sup- 
pose we  had  a  floor  carrying  a  live  load  of  150  per  sq.  ft. 
supported  on  beams  6  ft.  6  in.  apart.  The  total  load  on  a 
strip  12  ins.  wide  would  be  6^X150  =  975  plus  the  weight  of 


182  STRENGTH  OF  MATERIALS 

the  slab  itself.  This  will  probably  be  about  3|  ins.  thick, 
which  with  the  top  floor  will  cause  a  dead  load  of  say  45  Ibs. 
per  sq.  ft.  6|X45  =  292.5  Ibs.,  say  300  Ibs.  W  then  =  1275 
Ibs. 

„         .,.    ,,     WL     1275X6.5X12     _._  .     „ 
From  this  M  =  -^  =  -     ~fn~    ~  =  9945  in.-lbs. 

Since  6=12  and  M=107.46<i2,  we  can  say 


__ 
107.4X12 

from  which  d  =  2.78. 

This  value  of  d  is  knoTvn  as  the  effective  depth  and  we  must 
add  about  f-in.  of  concrete  below  the  steel  which  will  make 
the  slab  3J  ins.  thick  as  assumed. 

The  amount  of  steel  to  use  in  the  slab  is  found  by 

As  =  pbd 

=  .0077X12X2.78 
=  .265. 

This  means  that  we  must  use  .265  sq.  in.  of  steel  in  each  foot 
of  width  of  slab,  and  this  can  be  made  up  by  using  a  mesh  or 
rods.  If  a  mesh  reinforcement  is  used  the  area,  size,  etc., 
can  be  obtained  from  the  manufacturers'  catalog.  If  rods 
are  used  the  spacing  is  figured  so  that  the  correct  amount  of 
steel  shall  be  placed  in  each  foot  of  width.  It  is  customary 
to  use  rods  from  J-in.  to  J-in.  in  size  and  vary  the  spacing  to 
suit  the  condition.  For  example,  for  the  above  slab  if  we 
decide  to  use  f-in.  square  bars,  the  area  of  each  is  .14  sq.  in. 

265 

and  we  will  need  L   -7-  =  1.89  rods  in  each  12  ins.  of  slab. 
.14 

12 

This  would  mean  that  the  rods  would  be  placed  —^  =  6.35 

J-  .  ob/ 

say  6f  ins.  apart.  Rods  in  slabs  should  not  be  spaced  more 
than  2J  times  the  slab  thickness  apart  nor  nearer  together 


REINFORCED  CONCRETE  183 

than  2J  times  the  diameter  of  rod.  If  6|  ins.  is  considered 
as  too  wide  spacing  for  the  rods  we  can  use  i%-in.  rods  which 
have  an  area  of  .098.  The  spacing  can  be  found  directly  by 

12  X.  098 

265      =4.43,  say  4J-m.  ctrs. 

Rectangular  beams  are  designed  in  just  the  same  way  as 
slabs  with  the  exception  that  they  are  now  always  12  ins. 
wide  (they  should  not  be  less  than  6  ins.  wide),  and  the  steel 
area  cannot  always  be  made  up  as  near  to  the  correct  value 
as  with  slabs.  The  procedure  is  to  choose  a  number  of  rods 
which  will  give  an  area  as  near  as  possible  to  that  required 
and  then  investigate  for  Ss  and  Sc  using  the  formulae 


Ss  =  ~j — r%    and    Sc  = 

remembering  that 

As 


Tee  Beams.  In  a  floor  system  of  all  concrete  construc- 
tion, it  is  customary  to  cast  the  beams  and  slabs  at  one 
operation.  When  this  is  done  the  slab  can  be  considered  as 
part  of  the  beam  and  figured  as  part  of  the  compressive  area. 
Fig.  102  shows  a  diagram  of  a  tee  beam  with  the  slab 
forming  the  flange  of  the  beam.  Notice  that  the  width  b  is 
measured  on  the  flange  and  the  width  of  the  stem  is  called 
b1,  the  slab  thickness  is  t,  and  the  compressive  stress  is  con- 
sidered to  act  at  a  distance  z  below  the  top  of  the  flange. 

The  width  b  is  limited  and  is  usually  taken  as  about  12  or 

•  i    i  ,  i  •    •  . ,     span  of  the  beam 

14  times  t  provided  this  is  not  more  than  the  - 

or  more  than  the  distance  between  beams.  The  width  b1  is 
usually  about  2Xt  and  must  be  wide  enough  to  permit  the 
proper  placing  of  the  steel  rods,  which  should  not  be  spaced 


184  STRENGTH  OF  MATERIALS 

nearer  together  than  2J  diameter.  They  do  not,  however, 
have  to  be  all  in  one  row,  but  may  be  in  two  or  more  layers. 
The  value  of  d  can  be  assumed  or  can  be  approximated  by 
the  formula 

V+75blt 
15061    ; 
where 

V  —  Maximum  shear 

=  End  reaction. 

Then  an  approximate  value  of  As  can  be  obtained  by  the 
formula 


(H7) 


This  will  usually  give  an  area  of  steel  a  little  larger  than  that 
actually  required,  and  for  this  reason  some  engineers  consider 
the  beams  to  be  designed  at  this  point.  It  is  not  always  safe 
to  assume  that  the  stresses  will  all  be  within  proper  limits 
and,  therefore,  the  following  procedure  should  be  carried  out. 
It  should  be  noted  that  this  method  neglects  the  compressior 
which  exists  in  the  stem  of  the  beam  below  the  slab. 

Having  found  As  by  Formula  (117)  we  can  find  kd  by  the 
formula 

,,     2ndAs+bl2 


If  this  gives  a  value  of  kd  which  is  less  than  t,  then  the  beam  is 
not  considered  to  be  a  tee  beam,  but  it  is  figured  as  a  rect- 
angular beam.  If,  however,  the  value  of  kd  is  equal  to  more 
than  t,  we  proceed  as  follows:  Find  Z  by  the  formula 

3kd-2t     i 
~  2kd-t  X3' 
and  jd,  by  the  formula 

jd  =  d-z.       .......     (120) 


REINFORCED  CONCRETE  185 

Then  Sa  can  be  found  by 


_ 
Asjd' 

and  Sc  by  the  formula 


Mkd 


If  the  values  of  Ss  and  Se  are  not  safe,  new  values  must  be 
taken  for  b  and  d  and  the  calculations  made  again. 

The  design  of  a  tee  beam  is  best  illustrated  by  an  example, 
as  follows:  Suppose  the  beams  supporting  the  slab  pre- 
viously designed  were  18  ft.  long.  Find  their  proper  size 
and  amount  of  steel  reinforcing. 

The  total  load  supported  by  the  beam  will  be  6.5X18X 
(150+45)  plus  the  weight  of  the  stem  of  the  beam.  The 
stem  will  probably  be  about  8  ins.  wide  and  10  ins.  deep 
below  slab.  It  will  weigh  8X10  =  80  Ibs.  per  lin.  ft.  or 
18X80  =  1440  Ibs.  This  gives  a  total  load  of  22800  lbs.+ 

WL 
1440  Ibs.  =  24240  Ibs.  and  we  will  use  M  =  ~-~  again,  as  we 

will  bend  part  of  the  rods  up  at  the  ends  of  the  beam  and 
carry  them  over  into  the  next  beam.  M,  therefore,  is 


Figuring  for  a  value  of  d  we  get 

7+75&1*     12120+75X8X3.5 
d  =  -T506^  150X8  =11-85,  say  12ms. 

Checking  our  assumed  value  of  10  ins.  for  depth  of  stem 
we  get,  12  ins.  +  lj  ins.  (fireproofing  below  rods  which  is  not 
considered  as  part  of  the  beam)  =  13  J  ins.—  3J  ins.  (thick- 
ness of  slab)  =  10  ins. 


186  STRENGTH  OF  MATERIALS 

Next  figuring  for  As  by  Formula  (117) 

M  523580 

*~~7       A~~  ~7        3  —  f\\  -o.i 

(<*-|)&     (12-^  )X  16000 

say,  three  1-in.  sq.  rods. 


7^_,_..  . 

"  ~        2X15X3+2X42X3.5 


(NOTE.  —  The  value  of  6  was  used  as  12X0- 
This  value  of  k  means  that  we  have  a  true  tee  beam  since 
the  neutral  axis  is  below  the  flange. 
Then  finding  Z  by  Formula  (119) 

„    3kd-2tt     3X4.27-2X3.5  . 


2X4.72-3.5 
-z=  12-  1.15  =  10.85; 
M        523580 


*5' 


which  is  near  enough  16,000  to  pass,  and 


Mkd  523580X4.27 

42X3.  5(4.  27-1.  75)10.  85~55b; 


which  is  well  below  650,  the  allowable  value  of  Sc.  If  the 
value  of  Sc  found  by  the  formula  is  considered  too  small  we 
can  assume  either  a  smaller  value  of  b  or  smaller  value  of  d 
and  refigure  the  beam,  also  if  the  value  of  Sc  found  by  the 
formula  is  too  large  we  must  take  a  larger  value  of  d  or  b  or 
both  and  try  again. 


REINFORCED  CONCRETE 


187 


Stirrups.  Practically  every  beam  has  a  tendency  to 
fail  as  shown  in  Fig.  103,  and  to  prevent  this  the  beam 
should  be  provided  with  shear  rods  or  stirrups,  these  are 
placed  in  the  beam  as  indicated  in  Fig.  103  and  the  spacing 


"Not  more  than  "d" 
2  or  3  spaces 

FIG.  103. 


there  is  for  a  uniform  load.     The  stirrups  are  usually  made 
of  J-in.  to  J-in.  square  bars. 


ART.  64.    COLUMNS 

The  usual  method  of  reinforcing  concrete  columns  is 
shown  in  Fig.  100.  There  are  various  formulae  by  which 
the  safe  load  on  a  reinforced  concrete  column  can  be  figured 
but  they  are  seldom  used.  The  reason  is  that  the  principal 
object  of  the  steel  is  to  increase  the  strength  of  the  concrete, 
rather  than  to  support  much  of  the  load  itself.  This  in- 
creased value  of  the  compressive  strength  of  the  concrete 
varies  with  the  amount  of  steel  which  is  used,  and  some  of  the 
rules  governing  this  will  be  stated  and  illustrated.  One  rule 
is  that  if  from  1  to  4  per  cent  of  vertical  reinforcement 
together  with  about  1  per  cent  of  spirals  or  hoops  are  used, 
the  unit  stress  in  the  concrete  can  be  increased  45  per  cent 
over  the  strength  of  plain  concrete  and  the  steel  can  be  fig- 
ured at  7000  per  sq.  in.  of  section.  The  amount  of  reinforce- 
ment used  varies  with  the  load  on  the  column,  but  it  is  not 
customary  to  use  more  than  4  per  cent.  To  illustrate  how  the 
calculations  are  made,  suppose  we  had  a  load  of  350,000 


188  STRENGTH  OF  MATERIALS 

to  support  by  a  round  reinforced  concrete  column  12  ft. 
long.  (The  length  of  a  concrete  column  should  not  be  more 
than  15  times  the  diameter  of  the  column.)  An  approximate 
area  of  the  column  can  be  found  by  dividing  the  load  by 
about  700,  assuming  that  we  will  use  2J  per  cent  of  vertical 
reinforcement,  and  that  the  value  of  Ss  for  plain  concrete  is 
450,  the  value  Sc  for  the  reinforced  concrete  will  be  about  650. 
Therefore  350,000  divided  by  700  =  500.  This  would  mean 
a  column  24  ins.  in  diameter.  The  area  of  this  column  is 
455  sq.  in.  2J  per  cent  of  this  is  11.35  sq.  in.,  so  we  will  use 
eleven  1-in.  square  rods.  The  strength  of  the  column  is  found 
to  be  455X650+11X7000  =  373,000,  which  is  well  over  the 
required  strength.  The  amount  of  spiral  necessary  is  found 
as  follows :  If  we  use  1  per  cent  we  will  need  1  per  cent  of  the 
volume  of  the  column  as  spiral.  The  volume  of  a  section  of 
column  1  ft.  long  is  5428.8  cu.;  in.  1  per  cent  of  this  is  54.3 
cu.  in.  This  means  that  we  must  have  54.3  cu.  in.  of  spiral  in 
each  foot  of  length  of  the  column.  If  we  use  f-in.  round  rod 
for  spiral  there  will  be  75.4 X.I  1  =  8.29  cu.  in.  in  each  turn 

54  3 
around  the  column.     There  will  be  -   —=6.55  turns  around 

o .  yy 

the  column  per  foot  of  length,  or  we  could  say  that  the  pitch 

12 

of  the  spiral  will  be  7r—  =  1.83,  say  1 J  in. 
o.oo 

It  is  necessary  to  have  a  protective  coating  of  at  least  If 
ins.  outside  the  steel  which  cannot  be  figured  as  adding  any 
strength  to  the  column.  This  will  make  the  column  27  ins. 
in  diameter. 


ART.  65.     PRACTICAL  REVIEW  PROBLEMS 

PROB.  193.  Design  a  floor  slab,  using  the  critical  amount  of 
reinforcement  to  carry  a  load  of  200  Ibs.  per  sq.  ft.  on  a  span  of 
6  ft.  6  ins.  The  load  includes  the  weight  of  the  slab.  The  slab  is 
not  part  of  a  floor  and  must  be  considered  as  a  simple  beam. 


REINFORCED  CONCRETE  189 

PROB.  194.  A  rectangular  reinforced  concrete  beam  supports 
a  load  of  2500  Ibs.  per  ft.,  including  its  own  weight,  over  a  span  of 
12  ft.  Assuming  that  6  =  10  ins.,  d  =  22  ins.  andAs  =  2.5sq.  ins., 
what  stress  is  produced  in  the  concrete  and  in  the  steel? 

PROB.  195.  Design,  a  rectangular  reinforced  concrete  beam  to 
support  a  load  of  1500  Ibs.  per  foot,  in  addition  to  its  own  weight, 
on  a  span  of  16  ft.  Design  to  the  nearest  half  inch  and  assume 
6  =  12  ins.  Find  d,  total  depth  and  area  of  steel  for  main  reinforcing. 

PROB.  196.  Design  a  slab  to  support  a  load  of  250  Ibs.  per 
sq.  ft.,  in  addition  to  its  own  weight,  on  a  span  of  8  ft.  Slab  is 
part  of  a  floor  and,  therefore,  the  rods  can  be  bent  up  and  carried 
over  into  the  next  span. 

PROB.  197.  A  tee  beam  has  a  flange  3^  ins.  thick  and  supports 
a  load  of  1800  Ibs.  per  lin.  ft.,  which  includes  its  own  weight,  on  a 
span  of  15  ft.  The  value  of  d  is  14  ins.  and  6  =40  ins.  Find  how 
much  steel  is  required  and  the  values  of  the  stress  in  the  concrete 
and  the  steel.  Consider  beam  as  a  simple  beam. 

PROB.  198.  Design  a  tee  beam  on  a  span  of  20  ft.  to  sup- 
port a  load  of  2200  Ibs.  per  lin.  ft.,  which  includes  its  own  weight. 
The  flange  is  4  ins.  thick,  and  part  of  the  main  reinforcing  rods  are 
bent  up  and  carried  over  into  the  next  beam.  Take  bl  =  12  ins. 

PROB.  199.  Design  a  round  reinforced  concrete  column  to 
support  a  load  of  325,000  Ibs.  Use  3  per  cent  of  vertical  rods  and 
1  per  cent  of  spirals. 

PROB.  200.  Design  a  square  reinforced  concrete  column  to 
support  a  load  of  550,000  Ibs.,  using  4  per  cent  of  vertical  steel  rods 
and  1  per  cent  of  spirals. 

PROB.  201.  Prepare  a  schedule  of  spacing  for  stirrups  for  beam 
in  P">b.  198, 


INDEX 


A.S.M.E.    rules    for    riveted 

joints 141 

B 

Bars,  reinforcing 172,  173 

Beams,  cantilever 42 

— ,  cast-iron 81 

— ,  deflection  of 96 

— ,  simple 42 

—  steel 78 

Bending  moment 45 

general  equation 50 

diagrams 46,  47 

Beam  reactions 42 

Butt  joints 142 


Cast  iron 28 

-pipe 136 

Cement 169 

Center  of  gravity 57 

Coefficient  of  expansion 164 

Columns,  cast-iron 125 

— ,  concrete 175 

— ,  definition  of 118 

— ,  structural  steel 128 

— ,  types 119 

—  wood 124 


Combined  stresses,  compres- 
sion and  flexure ....   156 

,  tension  and  flexure .  .  .    151 

Component 2 

Concrete 169 

Crane  hooks . .  .   151 


Dangerous  section  of  beams . .     45 

Design  of  floor  slabs 180 

Diameter  of  rivet  holes .  .      .   139 


Efficiency  of  riveted  joints.  .  140 

Elastic  limit 11 

—  resilience 159 

Elongation 25 

Equilibriant 4 

Equilibrium 6 

— ,  conditions  of 7 

Extensometer 23 

F 

Factor  of  safety 13 

,  apparent 14 

,  real 14 

Force,  definition 1 

— ,  unit  of .  .  1 


191 


192 


INDEX 


Gravity 1 

— ,  center  of 57 

— ,  calculations  for 60,  61 


Horse-power 109 


Internal     resisting    moment, 

beams 53 

,  shafts 105 


Lap  joints 138 

M 

Machine  frames 154 

Method  of  solving  problems .  18 

Modulus  of  elasticity 26,  93 

resilience 160 

—  rupture 88 

Moment 2 

—  arm 2 

— ,  measure  of 2 


Open-hearth  furnace 34 

Overlap  of  joints 139 


Parallelogram  law 5 

Permanent  set 11 

Pitch  of  riveted  joints 139 

Polar  moment  of  inertia ....  106 

Power 109 

Properties  of  cement 170 

concrete 171 

-  I-beams 80 

materials .  .  39 


R 

PAGE 

Radius  of  gyration 120 

Rankine  column  formula ....  124 

Reduction  of  area 25 

Resilience 159 

—  of  beams 162 

Resisting  moment,  torsion .  .  .  104 

Resultant  force 3 

Results  of  tension  tests .  .  25 


Safe  loads  for  beams 71 

—  working  boiler  pressure.  .  .  148 

Section  modulus 56 

Shear 16 

—  diagrams 74,  75,  76 

Shrink  fits 166 

Steel 32 

— ,  Bessemer 33 

— ,  crucible. 32 

— ,  open-hearth 33 

Stirrups 187 

Stove 37 

Stress 9 

—  due  to  temperature  ....*..  164 


Tables,  coefficients  of  expan- 
sion    165 

— ,  constants  for  columns.  .  .  124 

— ,  elastic  limits 12 

— ,  factors  of  safety 15 

— ,  horse-power  of  shafting.  .  115 

— ,  modulus  of  elasticity ....  95 

— ,  —    —  rupture 89 

— ,  strength  of  steel  plates  .  .  35 

— ,  -    —  stone 38 

— ,  ultimate  shearing  strength  16 

— ,  unit  working  stresses ....  73 

— ,  weights  of  materials 40 

Tee  beams,  concrete 183 

Testing  machines 21 


INDEX 


193 


PAGE 

Tests,  compression 26 

— ,  tension 24 

— ,  transverse 89 

Thickness  of  boiler  shell 148 

-  pipe 136 

Timber 35 

Twisting  moment 102 


U 


Ultimate  resilience.  .  .    159 


Ultimate  strength 13,  25 

Unit  stress 9 

—  working  stress 14 


Vertical  shear 74 

W 

Work ' 109 

Wrought  iron 30 


THIS  BOOK  IS  DUE  ON  THE  LAST  DATE 
STAMPED  BELOW 


AN     INITIAL     FINE     OF    25      CENTS 

WILL  BE  ASSESSED  FOR  FAILURE  TO  RETURN 
THIS  BOOK  ON  THE  DATE  DUE.  THE  PENALTY 
WILL  INCREASE  TO  SO  CENTS  ON  THE  FOURTH 
DAY  AND  TO  $1.OO  ON  THE  SEVENTH  DAY 
OVERDUE. 


SEP  15 
SEP    1619& 


FEB 


26  1934 


1936 


Mi  ^ 


OCT  17  1941  M. 


FEB  16  1&43 

SEP    4  1943 


LD  21-20w-6,'32 


